{"id":7517,"date":"2024-01-20T11:02:27","date_gmt":"2024-01-20T05:32:27","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=7517"},"modified":"2024-01-22T17:50:20","modified_gmt":"2024-01-22T12:20:20","slug":"ts-inter-2nd-year-maths-2b-solutions-chapter-1-ex-1e","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-2nd-year-maths-2b-solutions-chapter-1-ex-1e\/","title":{"rendered":"TS Inter 2nd Year Maths 2B Solutions Chapter 1 Circle Ex 1(e)"},"content":{"rendered":"

Students must practice this TS Intermediate Maths 2B Solutions<\/a> Chapter 1 Circle Ex 1(e) to find a better approach to solving the problems.<\/p>\n

TS Inter 2nd Year Maths 2B Solutions Chapter 1 Circle Exercise 1(e)<\/h2>\n

I.<\/span><\/p>\n

Question 1.
\nDiscuss the relative position of the following pair of circles.
\n(i) x2<\/sup> + y2<\/sup> – 4x – 6y – 12 = 0, x2<\/sup> + y2<\/sup> + 6x + 18y + 26 = 0
\nSolution:
\nLet S \u2261 x2<\/sup> + y2<\/sup> – 4x – 6y – 12 = 0 ……..(1) and S’ \u2261 x2<\/sup> + y2<\/sup> + 6x + 18y + 26 = 0 ……..(2) be the given circles.
\nThe Centre of the circle (1) is C1<\/sub> = (2, 3)
\nCentre of circle (2) is C2<\/sub> = (-3, -9)
\nRadius of circle (1) is \\(\\sqrt{4+9+12}\\) = 5 = r1<\/sub>
\nRadius of circle (2) is \\(\\sqrt{9+81-26}\\) = 8 = r2<\/sub>
\nNow C1<\/sub>C2<\/sub> = \\(\\sqrt{(2+3)^2+(3+9)^2}\\) = 13
\nand r1<\/sub> + r2<\/sub> = 5 + 8 = 13
\nSince C1<\/sub>C2<\/sub> = r1<\/sub> + r2<\/sub>, the two circles touched each other externally.<\/p>\n

(ii) x2<\/sup> + y2<\/sup> + 6x + 6y + 14 = 0, x2<\/sup> + y2<\/sup> – 2x – 4y – 4 = 0
\nSolution:
\nTaking S \u2261 x2<\/sup> + y2<\/sup> + 6x + 6y + 14 = 0 ……….(1) and S’ \u2261 x2<\/sup> + y2<\/sup> – 2x – 4y – 4 = 0 ………(2)
\nThe Centre of the circle (1) is C1<\/sub>1 = (-3, -3)
\nand the Centre of the circle (2) is C2<\/sub> = (1, 2)
\nRadius of circle (1) is r1<\/sub> = \\(\\sqrt{9+9-14}\\) = 2
\nRadius of circle (2) is r2<\/sub> = \\(\\sqrt{1+4+4}\\) = 3
\nNow C1<\/sub>C2<\/sub> = \\(\\sqrt{(1+3)^2+(2+3)^2}\\) = \u221a41
\nr1<\/sub> + r2<\/sub> = 5
\nSince C1<\/sub>C2<\/sub> > r1<\/sub> + r2<\/sub>, each circle lies on the exterior of the other circle.<\/p>\n

(iii) (x – 2)2<\/sup> + (y + 1)2<\/sup> = 9, (x + 1)2<\/sup> + (y – 3)2<\/sup> = 4
\nSolution:
\nThe equation of the given circles are
\n(x – 2)2<\/sup> + (y + 1)2<\/sup> = 9 ………(1)
\n(x + 1)2<\/sup> + (y – 3)2<\/sup> = 4 ……….(2)
\nCentre of circle (1) is C1<\/sub> = (2, -1)
\nThe Centre of the circle (2) is C2<\/sub> = (-1, 3)
\nThe radius of the circle (1) is r1<\/sub> = 3
\nThe radius of the circle (2) is r2<\/sub> = 2
\nC1<\/sub>C2<\/sub> = \\(\\sqrt{(2+1)^2+(-1-3)^2}\\) = 5
\nand r1<\/sub> + r2<\/sub> = 3 + 2 = 5
\n\u2234 Since C1<\/sub>C2<\/sub> = r1<\/sub> + r2<\/sub>, the two circles touch each other externally.<\/p>\n

(iv) x2<\/sup> + y2<\/sup> – 2x + 4y – 4 = 0, x2<\/sup> + y2<\/sup> + 4x – 6y – 3 = 0
\nSolution:
\nGiven equations of circles are
\nx2<\/sup> + y2<\/sup> – 2x + 4y – 4 = 0 ……….(1)
\nx2<\/sup> + y2<\/sup> + 4x – 6y – 3 = 0 ……….(2)
\nCentre of circle (1) is C1<\/sub> = (1, -2)
\nThe Centre of the circle (2) is C2<\/sub> = (-2, 3)
\nRadius of circle (1) is r1<\/sub> = \\(\\sqrt{1+4+4}\\) = 3
\nRadius of circle (2) is r2<\/sub> = \\(\\sqrt{4+9+3}\\) = 4
\nDistance between centres C1<\/sub>C2<\/sub> = \\(\\sqrt{(1+2)^2+(-2-3)^2}=\\sqrt{9+25}=\\sqrt{34}\\)
\nr1<\/sub> + r2<\/sub> = 3 + 4 = 7 and |r1<\/sub> – r2<\/sub>| = |3 – 4| = 1
\n\u2234 |r1<\/sub> – r2<\/sub>| < C1<\/sub>C2<\/sub> < r1<\/sub> + r2<\/sub>
\n\u2234 The two given circles (1) and (2) intersect each other in two points.<\/p>\n

\"TS<\/p>\n

Question 2.
\nFind the number of possible common tangents that exist for the following pairs of circles.
\n(i) x2<\/sup> + y2<\/sup> + 6x + 6y + 14 = 0, x2<\/sup> + y2<\/sup> – 2x – 4y – 4 = 0
\nSolution:
\nThe given equations of circles are
\nx2<\/sup> + y2<\/sup> + 6x + 6y + 14 = 0 ………(1)
\nand x2<\/sup> + y2<\/sup> – 2x – 4y – 4 = 0 ……..(2)
\nCentre of circle (1) is C1<\/sub> = (-3, -3)
\nThe Centre of the circle (2) is C2<\/sub> = (1, 2)
\nRadius of circle (1) is r1<\/sub> = \\(\\sqrt{9+9-14}\\) = 2
\nRadius of circle (2) is r2<\/sub> = \\(\\sqrt{1+4+4}\\) = 3
\nC1<\/sub>C2<\/sub> = \\(\\sqrt{(-3-1)^2+(-3-2)^2}=\\sqrt{16+25}\\) = \u221a41
\nand r1<\/sub> + r2<\/sub> = 2 + 3 = 5
\nSince C1<\/sub>C2<\/sub> > r1<\/sub> + r2<\/sub> the given circles do not intersect each other.
\nThe number of possible common tangents that can be drawn to the above circles is ‘4’.<\/p>\n

(ii) x2<\/sup> + y2<\/sup> – 4x – 2y + 1 = 0, x2<\/sup> + y2<\/sup> – 6x – 4y + 4 = 0
\nSolution:
\nThe given equations of circles are
\nx2<\/sup> + y2<\/sup> – 4x – 2y + 1 = 0 ………(1)
\nand x2<\/sup> + y2<\/sup> – 6x – 4y + 4 = 0 ……….(2)
\nThe Centre of the circle (1) is C1<\/sub> = (2, 1)
\nThe Centre of the circle (2) is C2<\/sub> = (3, 2)
\nRadius of circle (1) is r1<\/sub> = \\(\\sqrt{4+1-1}\\) = 2
\nRadius of circle (2) is r2<\/sub> = \\(\\sqrt{9+4-4}\\) = 3
\nC1<\/sub>C2<\/sub> = \\(\\sqrt{(2-3)^2+(1-2)^2}=\\sqrt{1+1}=\\sqrt{2}\\)
\nand r1<\/sub> + r2<\/sub> = 2 + 3 = 5
\n\u2234 |r1<\/sub> – r2<\/sub>| = |2 – 3| = 1
\n\u2234 |r1<\/sub> – r2<\/sub>| < C1<\/sub>C2<\/sub> < r1<\/sub> + r2<\/sub>
\n\u2234 The two given circles intersect each other in two points.
\nThe number of possible common tangents that can be drawn to the circles is ‘2’.<\/p>\n

(iii) x2<\/sup> + y2<\/sup> – 4x + 2y – 4 = 0, x2<\/sup> + y2<\/sup> + 2x – 6y + 6 = 0
\nSolution:
\nThe given equations of circles are
\nx2<\/sup> + y2<\/sup> – 4x + 2y – 4 = 0 ……….(1)
\nand x2<\/sup> + y2<\/sup> + 2x – 6y + 6 = 0 …………(2)
\nCentre of circle (1) is C1<\/sub> = (2, -1)
\nThe Centre of the circle (2) is C2<\/sub> = (-1, 3)
\nRadius of circle (1) is r1<\/sub> = \\(\\sqrt{4+1+4}\\) = 3
\nRadius of circle (2) is r2<\/sub> = \\(\\sqrt{1+9-6}\\) = 2
\nC1<\/sub>C2<\/sub> = \\(\\sqrt{(2+1)^2+(-1-3)^2}=\\sqrt{9+16}\\) = 5
\nand r1<\/sub> + r2<\/sub> = 3 + 2 = 5
\nSince C1<\/sub>C2<\/sub> = r1<\/sub> + r2<\/sub>, the two circles touch externally and the number of possible common tangents is ‘3’.<\/p>\n

(iv) x2<\/sup> + y2<\/sup> = 4, x2<\/sup> + y2<\/sup> – 6x – 8y + 16 = 0
\nSolution:
\nThe equations of given circles are x2<\/sup> + y2<\/sup> = 4 ………(1) and x2<\/sup> + y2<\/sup> – 6x – 8y + 16 = 0 ……….(2)
\nThe Centre of the circle (1) is C1<\/sub> = (0, 0)
\nThe Centre of the circle (2) is C2<\/sub> = (3, 4)
\nThe radius of the circle (1) is r1<\/sub> = 2
\nRadius of circle (2) is r2<\/sub> = \\(\\sqrt{9+16-16}\\) = 3
\nC1<\/sub>C2<\/sub> = \\(\\sqrt{9+16}\\) = 5 = r1<\/sub> + r2<\/sub>
\nHence the two circles touch externally and the number of possible common tangents is ‘3’.’<\/p>\n

(v) x2<\/sup> + y2<\/sup> + 4x – 6y – 3 = 0, x2<\/sup> + y2<\/sup> + 4x – 2y + 4 = 0
\nSolution:
\nThe equations of circles are
\nx2<\/sup> + y2<\/sup> + 4x – 6y – 3 = 0 ……….(1)
\nand x2<\/sup> + y2<\/sup> + 4x – 2y + 4 = 0 ……….(2)
\nThe Centre of the circle (1) is C1<\/sub> = (-2, 3)
\nThe Centre of the circle (2) is C2<\/sub> = (-2, 1)
\nAlso radius of circle are r1 = \\(\\sqrt{4+9+3}\\) = 4
\nRadius of circle (2) are r2 = \\(\\sqrt{4+1-4}\\) = 1
\nNow C1<\/sub>C2<\/sub> = \\(\\sqrt{(-2+2)^2+(3-1)^2}\\) = 4
\nand |r1<\/sub> – r2<\/sub>| = |4 – 1| = 3
\n\u2234 C1<\/sub>C2<\/sub> < |r1<\/sub> + r2<\/sub>|, one circle lies completely inside the other circle.
\n\u2234 The number of common tangents that can be drawn to the circles S = 0 and S’ = 0 is zero.<\/p>\n

\"TS<\/p>\n

Question 3.
\nFind the internal centre of similitude of the circles x2<\/sup> + y2<\/sup> + 6x – 2y + 1 = 0 and x2<\/sup> + y2<\/sup> – 2x – 6y + 9 = 0.
\nSolution:
\nEquations of given circles are
\nx2<\/sup> + y2<\/sup> + 6x – 2y + 1 = 0 ……….(1)
\nx2<\/sup> + y2<\/sup> – 2x – 6y + 9 = 0 ……….(2)
\nThe Centre of the circle (1) is C1<\/sub> = (-3, 1)
\nThe Centre of the circle (2) is C2<\/sub> = (1, 3)
\nAlso radius of circle (1) is r1<\/sub> = \\(\\sqrt{9+1-1}\\) = 3
\nradius of circle (2) is r2<\/sub> = \\(\\sqrt{1+9-9}\\) = 1
\nNow C1<\/sub>C2<\/sub> = \\(\\sqrt{(-3-1)^2+(1-3)^2}=\\sqrt{16+4}\\) = \u221a20
\nand r1<\/sub> + r2<\/sub> = 3 + 1 = 4
\nSince C1<\/sub>C2<\/sub> > r1<\/sub> + r2<\/sub>, the two circles do not touch or do not intersect each other but each circle lies in the exterior of the other.
\n\u2234 Number of common tangents = 4
\nBoth internal and external centres of similitude exist.
\n\"TS
\nLet A be the internal centre of similitude and r1<\/sub> : r2<\/sub> = 3 : 1.
\n\u2234 A divides \\(\\overline{\\mathrm{C}_1 \\mathrm{C}_2}\\) in the ratio 3 : 1 internally.
\n\"TS
\n\u2234 The internal centre of similitude is (0, \\(\\frac{5}{2}\\))<\/p>\n

Question 4.
\nFind the external centre of similitude for the circles x2<\/sup> + y2<\/sup> – 2x – 6y + 9 = 0, x2<\/sup> + y2<\/sup> = 4.
\nSolution:
\nThe equations of given circles are x2<\/sup> + y2<\/sup> – 2x – 6y + 9 = 0 ……..(1) and x2<\/sup> + y2<\/sup> = 4 ……..(2)
\nCentre of circles are C1<\/sub> = (1, 3) and C2<\/sub> = (0, 0)
\nRadius of circles are r1<\/sub> = \\(\\sqrt{1+9-9}\\) = 1 and r2<\/sub> = \u221a4 = 2
\nNow C1<\/sub>C2<\/sub> = \\(\\sqrt{1+9}=\\sqrt{10}\\) and r1<\/sub> + r2<\/sub> = 1 + 2 = 3
\nSince C1<\/sub>C2<\/sub> > r1<\/sub> + r2<\/sub>, each circle lies in the exterior of the other circle, and the number of common tangents = 4
\n\u2234 Both internal and external centres of similitudes do exist in this case.
\nC1<\/sub> = (1, 3) = (x1<\/sub>, y1<\/sub>) and C2<\/sub> = (0, 0) = (x2<\/sub>, y2<\/sub>)
\n\"TS
\n\u2234 The external centre of the similitude of the circle is (2, 6).<\/p>\n

II.<\/span><\/p>\n

Question 1.
\n(i) Show that the circles x2<\/sup> + y2<\/sup> – 6x – 2y + 1 = 0, x2<\/sup> + y2<\/sup> + 2x – 8y + 13 = 0 touch each other. Find the point of contact and the equation of the common tangent at their point of contact. (Mar. ’11, ’10, ’09)
\nSolution:
\nLet S = x2<\/sup> + y2<\/sup> – 6x – 2y + 1 = 0 and S’ = -x2<\/sup> + y2<\/sup> + 2x – 8y + 13 = 0 be the given circles.
\nThe centre of circles are C1<\/sub> = (3, 1) and C2<\/sub> = (-1, 4)
\nRadius of circles are r1<\/sub> = \\(\\sqrt{9+1-1}\\) = 3 and r2<\/sub> = \\(\\sqrt{1+16-13}\\) = 2
\nDistance between centres C1<\/sub>C2<\/sub> = \\(\\sqrt{(3+1)^2+(1-4)^2}=\\sqrt{16+9}\\) = 5 and r1<\/sub> + r2<\/sub> = 3 + 2 = 5
\nSince C1<\/sub>C2<\/sub> = r1<\/sub> + r2<\/sub>, the two circles touch externally at a point.
\n\"TS
\nLet P be the point of contact of circles such that r1<\/sub> : r2<\/sub> = 3 : 2.
\nSince P divides C1<\/sub>, C2<\/sub> internally in the ratio 3 : 2,
\nthe coordinates of the point of contact
\n\"TS
\nThe equation of common tangent is S – S’ = 0
\n\u21d2 x2<\/sup> + y2<\/sup> – 6x – 2y + 1 – x2<\/sup> – y2<\/sup> – 2x + 8y + 13 = 0
\n\u21d2 -8x + 6y – 12 = 0
\n\u21d2 4x – 3y + 6 = 0
\n\u2234 The equation of the common tangent at the point of contact is 4x – 3y + 6 = 0.<\/p>\n

(ii) Show that x2<\/sup> + y2<\/sup> – 6x – 9y + 13 = 0, x2<\/sup> + y2<\/sup> – 2x – 16y = 0 touch each other. Find the point of contact and the equation of the common tangent at their point of contact.
\nSolution:
\nGiven equations of circles are S = x2<\/sup> + y2<\/sup> – 6x – 9y + 13 = 0 and S’ = x2<\/sup> + y2<\/sup> – 2x – 16y = 0
\nCentre of circles are C1<\/sub> = (3, \\(\\frac{9}{2}\\)) and C2<\/sub> = (1, 8)
\n\"TS
\nThe point P divides C1<\/sub>C2<\/sub> externally in the ratio of their radius \\(\\frac{\\sqrt{65}}{2}\\) : \u221a65 = 1 : 2
\n\"TS
\nThe equation of the common tangent at P(5, 1) to the circles S = 0 and S’ = 0 is S – S’ = 0.
\n\u21d2 x2<\/sup> + y2<\/sup> – 6x – 9y + 13 – x2<\/sup> – y2<\/sup> + 2x + 16y = 0
\n\u21d2 -4x + 7y – 13 = 0
\n\u21d2 4x – 7y + 13 = 0
\n\u2234 The equation of the required common tangent is 4x – 7y + 13 = 0.<\/p>\n

\"TS<\/p>\n

Question 2.
\nFind the equation of the circle which touches the circle x2<\/sup> + y2<\/sup> – 2x – 4y – 20 = 0 externally at (5, 5) with radius ‘5’.
\nSolution:
\n\"TS
\nLet C1<\/sub> = (h, k) be the centre and r1 be the radius of the required circle.
\nGiven r1<\/sub> = 5
\nEquation of the given circle is x2<\/sup> + y2<\/sup> – 2x – 4y – 20 = 0 ……….(1)
\nCentre of the circle C2<\/sub> = (1, 2) and radius r2<\/sub> = \\(\\sqrt{1+4+20}\\) = 5
\n\u2234 r1<\/sub> : r2<\/sub> = 5 : 5 = 1 : 1
\nPoint P(5, 5) divides C1<\/sub>C2<\/sub> in the ratio 1 : 1 internally.
\n\u2234 \\(\\left(\\frac{\\mathrm{h}+1}{2}, \\frac{\\mathrm{k}+2}{2}\\right)\\) = (5, 5)
\n\u21d2 \\(\\frac{\\mathrm{h}+1}{2}\\) = 5 and \\(\\frac{\\mathrm{k}+2}{2}\\) = 5
\n\u21d2 h = 9 and k = 18
\n\u2234 C1 = (h, k) = (9, 8)
\n\u2234 The equation of a required circle with centre (9, 8) and radius ‘5’ is (x – 9)2<\/sup> + (y – 8)2<\/sup> = 25
\n\u21d2 x2<\/sup> + y2<\/sup> – 18x – 16y + 120 = 0<\/p>\n

Question 3.
\nFind the direct common tangents of the circles x2<\/sup> + y2<\/sup> + 22x – 4y – 100 = 0 and x2<\/sup> + y2<\/sup> – 22x + 4y + 100 = 0. (New Model Paper)
\nSolution:
\nLet S = x2<\/sup> + y2<\/sup> + 22x – 4y – 100 = 0 ……….(1) and S’ = x2<\/sup> + y2<\/sup> – 22x + 4y + 100 = 0 …….(2) be the given circles.
\nThen centres of the circles are C1<\/sub> = (-11, +2) and C2<\/sub> = (11, -2)
\nThe radius of the circles are
\n\"TS
\nand r1<\/sub> + r2<\/sub> = 15 + 5 = 20
\nSince C1<\/sub>C2<\/sub> > r1<\/sub> + r2<\/sub>, the two circles are such that each lies on the exterior of the other.
\nHence both internal and external centre of similitude exists.
\n\"TS
\nNow r1<\/sub> : r2<\/sub> = 15 : 5 = 3 : 1
\nPoint P divides C1<\/sub>, C2<\/sub> externally in the ratio of radii 3 : 1.
\n\"TS
\nThe equation of direct common tangent passing through (22, -4) having slope ‘m’ is y + 4 = m(x – 22)
\n\u21d2 mx – y – 4 – 22m = 0 …….(1)
\nSince the line is tangent to S’ = 0.
\nPerpendicular distance from C2<\/sub> = (11, -2) to the line (1) = radius of the circle S’ = 0
\n\u2234 \\(\\left|\\frac{m(11)+2-4-22 m}{\\sqrt{m^2+1}}\\right|\\) = 5
\n\u21d2 \\(\\left|\\frac{-(11 m+2)}{\\sqrt{m^2+1}}\\right|\\) = 5
\n\u21d2 (11m + 2)2<\/sup> = 25(m2<\/sup> + 1)
\n\u21d2 121m2<\/sup> + 44m + 4 = 25m2<\/sup> + 25
\n\u21d2 96m2<\/sup> + 44m – 21 = 0
\n\u21d2 96m2<\/sup> + 72m – 28 m – 21 = 0
\n\u21d2 24m(4m + 3) – 7(4m + 3) = 0
\n\u21d2 (24m – 7) (4m + 3) = 0
\n\u21d2 m = \\(\\frac{-3}{4}\\) or m = \\(\\frac{7}{24}\\)
\n\u2234 Equations of direct common tangents from (1) are y + 4 = \\(\\frac{-3}{4}\\)(x – 22)
\n\u21d2 4y + 16 = -3x + 66
\n\u21d2 3x + 4y – 50 = 0
\nand y + 4 = \\(\\frac{7}{24}\\)(x – 22)
\n\u21d2 24y + 96 = 7x – 154
\n\u21d2 7x – 24y – 250 = 0
\n\u2234 The equations of direct common tangents are given by 3x + 4y – 50 = 0 and 7x – 24y – 250 = 0.<\/p>\n

\"TS<\/p>\n

Question 4.
\nFind the transverse common tangents of the circles x2<\/sup> + y2<\/sup> – 4x – 10y + 28 = 0 and x2<\/sup> + y2<\/sup> + 4x – 6y + 4 = 0.
\nSolution:
\nLet S = x2<\/sup> + y2 – 4x – 10y + 28 = 0 ………(1)
\nS’ = x2<\/sup> + y2 + 4x – 6y + 4 = 0 ……….(2)
\nbe the given circles.
\nCentres of the circles are C1<\/sub> = (2, 5) and C2<\/sub> = (-2, 3)
\nAlso, the radius of circles are r1<\/sub> = \\(\\sqrt{4+25-28}\\) = 1 and r2<\/sub> = \\(\\sqrt{4+9-4}\\) = 3
\nAlso C1<\/sub>C2<\/sub> = \\(\\sqrt{(2+2)^2+(5-3)^2}\\) = \u221a20 = 2\u221a5 and r1<\/sub> + r2<\/sub> = 4
\nSince C1<\/sub>C2<\/sub> > r1<\/sub> + r2<\/sub>, we have each circle lying in the exterior of other circles.
\nHence we get 2 direct common tangents and 2 transverse common tangents that can be drawn.
\n\"TS
\nPoint P divides C1<\/sub>C2<\/sub> internally in the ratio r1<\/sub> : r2<\/sub> = 1 : 3.
\nLet P be the internal centre of similitude.
\n\"TS
\nLet the equation of transverse common tangent be y – \\(\\frac{9}{2}\\) = m(x – 1) ……(1)
\n\u21d2 2y – 9 = 2mx – 2m
\n\u21d2 2mx – 2y + (9 – 2m) = 0 ……….(2)
\nSince line (2) is a tangent to the circle S = 0, the perpendicular distance from C1<\/sub> (2, 5) to line (2) is equal to the radius of the circle S = 0.
\n\u2234 \\(\\left|\\frac{2 m(2)-2(5)+9-2 m}{\\sqrt{4 m^2+4}}\\right|\\) = 1
\n\u21d2 |2m – 1| = \\(2 \\sqrt{m^2+1}\\)
\n\u21d2 4m2<\/sup> – 4m + 1 = 4(m2<\/sup> + 1)
\n\u21d2 4m2<\/sup> – 4m + 1 – 4m2<\/sup> – 4 = 0
\n\u21d2 -4m – 3 = 0
\n\u21d2 m = \\(\\frac{-3}{4}\\)
\n\u2234 Equation of tangent is y – \\(\\frac{9}{2}\\) = \\(\\frac{-3}{4}\\)(x – 1)
\n\u21d2 \\(\\frac{2 y-9}{2}=-\\frac{3}{4}(x-1)\\)
\n\u21d2 2(2y – 9) = -3(x – 1)
\n\u21d2 4y – 18 = -3x + 3
\n\u21d2 3x + 4y – 21 = 0
\nSince the m2<\/sup> term is canceled, the slope of one of the transverse common tangents is not defined.
\nOne transverse common tangent is passing through P(1, \\(\\frac{9}{2}\\)) and parallel to y-axis.
\n\u2234 The equation of other transverse common tangents is x = 1.
\n\u2234 The equations of transverse common tangents are x = 1 and 3x + 4y – 21 = 0.<\/p>\n

Question 5.
\nFind the pair of tangents from (4, 10) to the circle x2<\/sup> + y2<\/sup> = 25.
\nSolution:
\nLet P(x1<\/sub>, y1<\/sub>) be the given point.
\nThe equation to the pair of tangents drawn from P(4, 10) to S = 0 is \\(\\mathrm{S}_1^2=\\mathrm{SS}_{11}\\).
\nGiven S = x2<\/sup> + y2<\/sup> – 25 = 0
\n(xx1<\/sub> + yy1<\/sub> – 25)2<\/sup> = (x2<\/sup> + y2<\/sup> – 25) (\\(x_1^2+y_1^2\\) – 25)
\n\u21d2 (4x + 10y – 25)2<\/sup> = (x2<\/sup> + y2<\/sup> – 25) (16 + 100 – 25)
\n\u21d2 (4x + 10y – 25)2<\/sup> = (x2<\/sup> + y2<\/sup> – 25) (91)
\n\u21d2 16x2<\/sup> + 100y2<\/sup> + 625 + 80xy – 500y – 200x = 91x2<\/sup> + 91y2<\/sup> – 2275
\n\u21d2 75x2<\/sup> – 9y2<\/sup> – 80xy + 200x + 500y – 2900 = 0<\/p>\n

\"TS<\/p>\n

Question 6.
\nFind the pair of tangents drawn from (0, 0) to x2<\/sup> + y2<\/sup> + 10x + 10y + 40 = 0.
\nSolution:
\nLet P(x1<\/sub>, y1<\/sub>) be the given point.
\nThe equation to the pair of tangents drawn from P(0, 0) to S = 0 is \\(\\mathrm{S}_1^2=\\mathrm{SS}_{11}\\).
\nGiven S = x2<\/sup> + y2<\/sup> + 10x + 10y + 40 = 0
\n\u2234 The equation to the pair of tangents from (0, 0) to S = 0 is [xx1<\/sub> + yy1<\/sub> + g(x + x1<\/sub>) + f(y + y1<\/sub>) + c]2<\/sup> = (x2<\/sup> + y2<\/sup> + 2gx + 2fy + c) (\\(x_1^2+y_1^2\\) + 2gx1<\/sub> + 2fy1<\/sub> + c)
\n\u21d2 (5x + 5y + 40)2<\/sup> = (x2<\/sup> + y2<\/sup> + 10x + 10y + 40) (40)
\n\u21d2 25x2<\/sup> + 25y2<\/sup> + 1600 + 50xy + 400xy + 400x = 40(x2<\/sup> + y2<\/sup> + 10x + 10y + 40)
\n\u21d2 15x2<\/sup> + 15y2<\/sup> – 50xy = 0
\n\u21d2 3x2<\/sup> + 3y2<\/sup> – 10xy = 0<\/p>\n

III.<\/span><\/p>\n

Question 1.
\nFind the equation of the circle which touches x2<\/sup> + y2<\/sup> – 4x + 6y – 12 = 0 at (-1, 1) internally with a radius of ‘2’.
\nSolution:
\nThe given circle is x2<\/sup> + y2<\/sup> – 4x + 6y – 12 = 0
\nThe centre of the circle is (2, -3) and radius = \\(\\sqrt{4+9+12}\\) = 5
\nLet the centre of the required circle be (h, k), and given that 2 is the radius with the point of contact of circles internally as (-1, 1).
\nSuppose (x1<\/sub>, y1<\/sub>) = (2, -3), (x2<\/sub>, y2<\/sub>) = (h, k), r1<\/sub> = 5, r2<\/sub> = 2
\npoint of contact (x, y) = (-1, 1)
\nusing the formula for the external division since two circles touch internally.
\n\"TS
\n\u21d2 (5x – 1)2<\/sup> + (5y – 3)2<\/sup> = 100
\n\u21d2 25x2<\/sup> + 25y2<\/sup> – 10x – 30y – 90 = 0
\n\u21d2 5x2<\/sup> + 5y2<\/sup> – 2x – 6y – 18 = 0<\/p>\n

Question 2.
\nFind all common tangents of the following pairs of circles.
\n(i) x2<\/sup> + y2<\/sup> = 9 and x2<\/sup> + y2<\/sup> – 16x + 2y + 49 = 0
\nSolution:
\nLet S = x2<\/sup> + y2<\/sup> – 9 = 0 and S’ = x2<\/sup> + y2<\/sup> – 16x + 2y + 49 = 0 be the given circles.
\nCentre of the circle S = 0 is C1<\/sub> (0, 0) and radius = 3
\nCentre of the circle S’ = 0 is C2<\/sub> (8, -1) and radius r2 = \\(\\sqrt{64+1-49}\\) = 4
\nr1<\/sub> + r2<\/sub> = 3 + 4 = 7 and C1<\/sub>C2<\/sub> = \\(\\sqrt{64+1}=\\sqrt{65}\\) > 7
\nSince C1<\/sub>C2<\/sub> > r1<\/sub> + r2<\/sub> we get each circle lies completely in the exterior of the other.
\n\u2234 Two direct and two transverse common tangents can be drawn to S = 0 and S’ = 0.
\nr1<\/sub> : r2<\/sub> = 3 : 4
\nLet A be the internal centre of similitude and A divides C1<\/sub>C2<\/sub> in the ratio 3 : 4 internally
\n\"TS
\n\u21d2 64x2<\/sup> + y2<\/sup> + 441 – 16xy + 42y – 336x = 16(x2<\/sup> + y2<\/sup> – 9)
\n\u21d2 48x2<\/sup> – I6xy – 15y2<\/sup> + 42y – 336x + 585 = 0
\n\u21d2 48x2<\/sup> – 36xy + 20xy – 15y2<\/sup> + 42y – 336x + 585 = 0
\n\u21d2 12x(4x – 3y) + 5y(4x – 3y) + 42y – 336x + 585 = 0
\n\u21d2 (12x + 5y)(4x – 3y) – 336x + 42y + 585 = 0
\nLet 48x2<\/sup> – 16xy – 15y2<\/sup> + 42y – 336x + 585 = (12x + 5y + l) (4x – 3y + m)
\n= 48x2<\/sup> – 36xy + 12mx + 20xy – 15y2<\/sup> + 5ym + 4lx + 3ly + lm
\n= 48x2<\/sup> – 16xy – 15y2<\/sup> + (4l + 12m)x + (-3l + 5m)y + lm
\nEquating coefficients of x, y, and the constant term
\n4l + 12m = -336
\n\u21d2 l + 3m = -84 ………(1)
\n-3l + 5m = 42
\n\u21d2 3l – 5m = -42 ……..(2)
\nand lm = 585 …….(3)
\nFrom (1), 5l + 15m = -420
\nFrom (2), 9l – 15m = -126
\n\u2234 14l = -546
\n\u21d2 l = -39
\nFrom (1), -39 + 3m = -84
\n\u21d2 3m = -45
\n\u21d2 m = -15
\n\u2234 48x2<\/sup> – 16xy – 15y2<\/sup> + 42y – 336x + 585 = (12x + 5y – 39) (4x – 3y – 15)
\n\u2234 The transverse common tangents are 12x + 5y – 39 = 0 and 4x – 3y – 15 = 0
\nLet B be the external centre of the similitude of circles.
\n\u2234 B divides C1C2 in the ratio 3 : 4 externally
\n\u2234 B = \\(\\left(\\frac{3(8)-4(0)}{3-4}, \\frac{3(-1)-4(0)}{3-4}\\right)\\) = (-24, 3)
\nThe equation is the pair of direct common tangents to the circle S = 0 from (-24, 3) is \\(S_1^2=S S_{11}\\)
\n\u21d2 [x(-24) + y(3) – 9]2<\/sup> = (x2<\/sup> + y2<\/sup> – 9) (576 + 9 – 9)
\n\u21d2 (-24x + 3y – 9)2<\/sup> = (576) (x2<\/sup> + y2 – 9)
\n\u21d2 9(-8x + y – 3)2<\/sup> = 9 \u00d7 64 (x2<\/sup> + y2 – 9)
\n\u21d2 (-8x + y – 3)2<\/sup> = 64(x2<\/sup> + y2 – 9)
\n\u21d2 64x2<\/sup> + y2<\/sup> + 9 – 16xy – 6y + 48x = 64x2<\/sup> + 64y2<\/sup> – 576
\n\u21d2 63y2<\/sup> + 16xy – 48x + 6y – 585 = 0
\n\u21d2 y(63y + 16x) – 48x + 6y – 585 = 0
\n\u2234 63y2<\/sup> + 16xy – 48x + 6y – 585 = (y + l) (63y + 16x + m)
\nEquating coefficients of x, y, and constants
\n16l = 48 ……..(4)
\n63l + m = 6 ………(5)
\nlm = -585 ……….(6)
\n\u2234 l = -3 and from (6), m = 195
\n\u2234 63y2<\/sup> + 16xy – 48x + 6y – 585 = (y – 3) (63y + 16x + 195)
\n\u2234 The equations of direct common tangents to the circles S = 0, S’ = 0 are y – 3 = 0, 63y + 16x + 195 = 0.
\nHence equations of all common tangents to the circles S = 0 and S’ = 0 are 4x – 3y – 15 = 0, 12x + 5y – 39 = 0 and y – 3 = 0, 16x + 63y + 195 = 0.<\/p>\n

(ii) x2<\/sup> + y2<\/sup> + 4x + 2y – 4 = 0 and x2<\/sup> + y2<\/sup> – 4x – 2y + 4 = 0
\nSolution:
\nLet S = x2<\/sup> + y2<\/sup> + 4x + 2y – 4 = 0 and S’ = x2<\/sup> + y2<\/sup> – 4x – 2y + 4 = 0
\nCentre of S = 0 is C1<\/sub> (-2, 1) and the radius is r1<\/sub> = \\(\\sqrt{4+1+4}\\) = 3
\nCentre of S’ = 0 is C2<\/sub> (2, 1) and the radius r2<\/sub> = \\(\\sqrt{4+1-4}\\) = 1
\nAlso C1<\/sub>C2<\/sub> = \\(\\sqrt{(2+2)^2+(1+1)^2}=\\sqrt{16+4}\\) = \u221a20 = 2\u221a5 and r1<\/sub> + r2<\/sub> = 3 + 1 = 4
\nSince C1<\/sub>C2<\/sub> > r1<\/sub> + r2<\/sub>, each circle lies in the exterior of the other circle.
\n\u2234 Two direct and two transverse common tangents can be drawn to the circles S = 0 and S’ = 0.
\nLet A, B be the internal and external centres of similitudes of circles S = 0 and S’ = 0
\nA divides \\(\\overline{\\mathrm{C}_1 \\mathrm{C}_2}\\) in the ratio r1<\/sub> : r2<\/sub> = 3 : 1 internally
\n\u2234 A = \\(\\left(\\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \\frac{m_1 y_2+m_2 y_1}{m_1+m_2}\\right)\\)
\n\"TS
\n\"TS
\n\u21d2 -(2x + y – 1)2<\/sup> = (x2<\/sup> + y2<\/sup> + 4x + 2y – 4)
\n\u21d2 (2x + y – 1)2<\/sup> = (x2<\/sup> + y2<\/sup> + 4x + 2y – 4)
\n\u21d2 4x2<\/sup> + y2<\/sup> + 1 + 4xy – 2y – 4x = x2<\/sup> + y2<\/sup> + 4x + 2y – 4
\n\u21d2 3x2<\/sup> + 4xy – 8x – 4y + 5 = 0
\n\u21d2 x(3x + 4y) – 8x – 4y + 5 = 0
\n\u21d2 3x2<\/sup> + 4xy – 8x – 4y + 5 = (x + l) (3x + 4y + m)
\nComparing the coefficients of x, y, and constants
\n3l + m = -8 ………(1)
\n4l = -4 ……….(2)
\nlm = 5 ……….(3)
\n\u2234 From (2), l = -1 and From (3), m = -5
\n\u2234 3x2<\/sup> + 4xy – 8x – 4y + 5 = (x – 1) (3x + 4y – 5)
\nThe equations of transverse common tangents are x – 1 = 0, 3x + 4y – 5 = 0.
\nThe equation of pair of direct common tangents to the circle S = 0 is \\(\\mathrm{s}_1^2=\\mathrm{S} \\mathrm{S}_{11}\\).
\n\u21d2 [x(4) + y(2) + 2(x + 4) + 1(y + 2) – 4]2<\/sup> = (x2<\/sup> + y2<\/sup> + 4x + – 2y – 4) (16 + 4 + 16 + 4 – 4)
\n\u21d2 (6x + 3y + 6)2<\/sup> = (x2<\/sup> + y2<\/sup> + 4x + 2y – 4)(36)
\n\u21d2 9(2x + y + 2)2<\/sup> = 36(x2<\/sup> + y2<\/sup> + 4x + 2y – 4)
\n\u21d2 (2x + y + 2)2<\/sup> = 4(x2<\/sup> + y2<\/sup> + 4x + 2y – 4)
\n\u21d2 4x2<\/sup> + y2<\/sup> + 4 + 4xy + 4y + 4x = 4x2<\/sup> + 4y2<\/sup> + 16x + 8y – 16
\n\u21d2 3y2<\/sup> – 4xy + 8x + 4y – 20 = 0
\n\u21d2 y(3y – 4x) + 8x + 4y – 20 = 0
\n\u2234 3y2<\/sup> – 4xy + 8x + 4y – 20 = (y + l) (3y – 4x + m)
\nComparing coefficients of x, y, and constants we get
\n-4l = 8 ………(4)
\n3l + m = 4 ………(5)
\nlm = -20 ………(6)
\nFrom (4), l = -2 and from (6), m = 10
\n\u2234 3y2<\/sup> – 4xy + 8x + 4y – 20 = (y – 2) (3y – 4x + 10)
\nHence the equations of direct common tangents are y – 2 = 0, 3y – 4x + 10 = 0.
\n\u2234 The equations of all common tangents to the circles S = 0, S’ = 0 are x – 1 = 0, 3x + 4y – 5 = 0 and y – 2 = 0, 4x – 3y – 10 = 0<\/p>\n

\"TS<\/p>\n

Question 3.
\nFind the pair of tangents drawn from (3, 2) to the circle x2<\/sup> + y2<\/sup> – 6x + 4y – 2 = 0.
\nSolution:
\nLet S \u2261 x2<\/sup> + y2<\/sup> – 6x + 4y – 2 = 0 be the given circle.
\nThe equation of pair of tangents drawn from (3, 2) to the circle S = 0 is \\(S_1^2=S . S_{11}\\)
\n\u21d2 [x(3) + y(2) – 3(x + 3) + 2(y + 2) – 2]2<\/sup> = (x2<\/sup> + y2<\/sup> – 6x + 4y – 2) (32<\/sup> + 22<\/sup> – 18 + 8 – 2)
\n\u21d2 (4y – 7)2<\/sup> = 1 . (x2<\/sup> + y2<\/sup> – 6x + 4y – 2)
\n\u21d2 16y2<\/sup> – 56y + 49 = x2<\/sup> + y2<\/sup> – 6x + 4y – 2
\n\u21d2 x2<\/sup> – 15y2<\/sup> – 6x + 60y – 51 = 0<\/p>\n

Question 4.
\nFind the pair of tangents drawn from (1, 3) to the circle x2<\/sup> + y2<\/sup> – 2x + 4y – 11 = 0 and also find the angle between them.
\nSolution:
\nDenote S \u2261 x2<\/sup> + y2<\/sup> – 2x + 4y – 11 = 0
\nThe equation of pair of tangents from (1, 3) to the circle S = 0 is by the formula \\(S_1^2=S . S_{11}\\)
\n\u21d2 [x(1) + y(3) – (x + 1) + 2(y + 3) – 11]2<\/sup> = (x2<\/sup> + y2<\/sup> – 2x + 4y – 11) (12<\/sup> + 32<\/sup> – 2 + 12 – 11)
\n\u21d2 (5y – 6)2<\/sup> = (x2<\/sup> + y2<\/sup> – 2x + 4y – 11) (9)
\n\u21d2 25y2<\/sup> – 60y + 36 = 9x2<\/sup> + 9y2<\/sup> – 18x + 36y – 99
\n\u21d2 9x2<\/sup> – 16y2<\/sup> – 18x + 96y – 135 = 0
\nIf \u03b8 is the angle between the above pair of tangents to S = 0, then
\n\"TS
\n(or) \u03b8 = \\(\\cos ^{-1}\\left(\\frac{7}{25}\\right)\\) is the angle between pair of tangents to S = 0.<\/p>\n

Question 5.
\nFind the pair of tangents from the origin to the circle x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0 and hence deduce a condition for these tangents to the perpendicular.
\nSolution:
\nLet S \u2261 x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0 be the given circle.
\nThen the equation of the pair of tangents drawn from (0, 0) to S = 0 is by the formula \\(S_1^2=S . S_{11}\\)
\n\u21d2 [x(0) + y(0) + g(x + 0) + f(y + 0) + c]2<\/sup> = (x2<\/sup> + y2<\/sup> + 2gx + 2fy + c) (c)
\n\u21d2 (gx + fy + c)2<\/sup> = cx2<\/sup> + cy2<\/sup> + 2gcx + 2fcy + c2<\/sup>
\n\u21d2 (g2<\/sup>x2<\/sup> + f2<\/sup>y2<\/sup> + c2<\/sup> + 2fgxy + 2fcy + 2cgx) = cx2<\/sup> + cy2<\/sup> + 2gcx + 2fcy + c2<\/sup>
\n\u21d2 x2<\/sup>(g2<\/sup> – c) + y2<\/sup>(f2<\/sup> – c) + 2fgxy = 0
\nGiven that pair of tangents represented by this equation are perpendicular we have
\ncoefficient of x2<\/sup> + coefficient of y2<\/sup> = 0
\n\u21d2 g2<\/sup> – c + f2<\/sup> – c = 0
\n\u21d2 g2<\/sup> + f2<\/sup> = 2c<\/p>\n

\"TS<\/p>\n

Question 6.
\nFrom a point on the circle x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0 two tangents are drawn to the circle x2<\/sup> + y2<\/sup> + 2gx + 2fy + c sin2<\/sup>\u03b1 + (g2<\/sup> + f2<\/sup>) cos2<\/sup>\u03b1 = 0 (0 < \u03b1 < \u03c0\/2). Prove that the angle between them is 2\u03b1.
\nSolution:
\nLet S = x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0 and S’ = x2<\/sup> + y2<\/sup> + 2gx + 2fy + c sin2<\/sup>\u03b1 + (g2<\/sup> + f2<\/sup>) cos2<\/sup>\u03b1 = 0 be the given circles.
\nLet P(x1<\/sub>, y1<\/sub>) be any point on S = 0 then
\n\"TS
\nLet \u03b8 be the angle between the tangents drawn from P(x1<\/sub>, y1<\/sub>) to S’ = 0 then
\n\"TS
\n= \\(\\frac{\\left(g^2+f^2-c\\right)\\left(\\cos ^2 \\alpha-\\sin ^2 \\alpha\\right)}{\\left(g^2+f^2-c\\right)\\left(\\cos ^2 \\alpha+\\sin ^2 \\alpha\\right)}\\)
\n= cos 2\u03b1
\n\u2234 cos \u03b8 = cos 2\u03b1 \u21d2 \u03b8 = 2\u03b1
\n\u2234 The angle between the tangents drawn from P(x1, y1) to the circle S’ = 0 is 2\u03b1.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice this TS Intermediate Maths 2B Solutions Chapter 1 Circle Ex 1(e) to find a better approach to solving the problems. TS Inter 2nd Year Maths 2B Solutions Chapter 1 Circle Exercise 1(e) I. Question 1. Discuss the relative position of the following pair of circles. (i) x2 + y2 – 4x – … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/7517"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=7517"}],"version-history":[{"count":3,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/7517\/revisions"}],"predecessor-version":[{"id":7550,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/7517\/revisions\/7550"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=7517"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=7517"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=7517"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}