TS Intermediate Maths 2B Solutions<\/a> Chapter 1 Circle Ex 1(e) to find a better approach to solving the problems.<\/p>\nTS Inter 2nd Year Maths 2B Solutions Chapter 1 Circle Exercise 1(e)<\/h2>\n
I.<\/span><\/p>\nQuestion 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<\/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<\/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
\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
\n\u2234 The internal centre of similitude is (0, \\(\\frac{5}{2}\\))<\/p>\nQuestion 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
\n\u2234 The external centre of the similitude of the circle is (2, 6).<\/p>\nII.<\/span><\/p>\nQuestion 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> + y