{"id":13141,"date":"2024-03-14T01:57:15","date_gmt":"2024-03-13T20:27:15","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=13141"},"modified":"2024-03-16T17:53:02","modified_gmt":"2024-03-16T12:23:02","slug":"ts-inter-2nd-year-maths-2b-circles-important-questions","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-2nd-year-maths-2b-circles-important-questions\/","title":{"rendered":"TS Inter 2nd Year Maths 2B Circles Important Questions"},"content":{"rendered":"

Students must practice these\u00a0TS Inter 2nd Year Maths 2B Important Questions<\/a> Chapter 1 Circles to help strengthen their preparations for exams.<\/p>\n

TS Inter 2nd Year Maths 2B\u00a0Circles Important Questions<\/h2>\n

Very Short Answer Type Questions\u00a0 \u00a0 \u00a0\u00a0<\/span><\/p>\n

Question 1.
\nFind the equation of circle with centre (1, 4) and radius 5.
\nSolution:
\nStandand equation of circle with centre
\n(h. k) and radius \u2018r\u2019 is (x – h)2<\/sup> + (y – k)2<\/sup> r2<\/sup>
\n(h, k) = (1, 4) and r= 5
\n\u2234 Equation of circle is (x – 1)2<\/sup> + (4)2<\/sup> = 25
\nx2<\/sup>. y2 <\/sup>-2x – 8y+ 17=25
\n= x2<\/sup>+y2<\/sup>– 2x – 8y – 8 = 0<\/p>\n

\"TS<\/p>\n

Question 2.
\nFind the centre and radius of the circle
\nx2<\/sup> + y2<\/sup> + 2x – 4y – 4 = 0.
\nSolution:
\nComparing with x2<\/sup>+y2<\/sup>+2gx+2fy+c = 0
\n2g=2, 21= – 4, and c = – 4
\n\u2234 Centre=(-g,-f)=(-1,2) and
\n\u2234 Radius = \\(\\sqrt{g^2+f^2-c}=\\sqrt{1+4+4}=3\\)<\/p>\n

Question 3.
\nFind the centre and radius of the circle
\n3x2<\/sup> + 3y2<\/sup> – 6x + 4y – 4 = 0.
\nSolution:
\nGiven equation can be written as
\n\"TS<\/p>\n

Question 4.
\nFind the equation of the circle whose centre is (-.1, 2) and which passes through (5,6).
\nSolution:
\nLet C (-1, 2) be the centre of the circle. Since (5, 6) is a point on the circle, the radius of the circle.
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 5.
\nFind the equation of circle passing through (2,3) and concentric with the circle
\nx2<\/sup> + y2<\/sup> + 8x + 12y + 15 = 0.
\nSolution:
\nLet the equation of required circle be
\nx2<\/sup> + y2<\/sup> + 2gx + 2fy + c\u2018 = 0
\nIf It passes through (2, 3) then
\n4+9+8(2)+ 12(3) + c\u2018 = 0
\n= 65 + c\u2019 = 0 \u21d4 c\u2019 = – 65
\n\u2234 The equation of required circle is x2<\/sup> + y2<\/sup> + 8x + 12y – 65 = 0.<\/p>\n

Question 6.
\nIf the circle x2<\/sup> + y2<\/sup> +ax+by -12 = 0 has the centre at (2,3) then find a, b and the radius of the circle.
\nSolution:
\nThe equation of circle is
\nx2<\/sup> + y2<\/sup> +ax+by -12 = 0
\nCentre 01 the circle = \\(\\left(\\frac{-\\mathrm{a}}{2},-\\frac{\\mathrm{b}}{2}\\right)\\) = (2, 3) (given)
\n\u2234 a =- 4 and b = – 6
\n\u2234 Radius of the circle = \\(\\sqrt{4+9+12}=5\\)<\/p>\n

Question 7.
\nIf the circle x2<\/sup> + y2 <\/sup>– 4x+6y+a = 0 has radius 4 then lead \u2018a\u2019.
\nSolution:
\nGiven equation of circle is
\nx2<\/sup> + y2 <\/sup>– 4x + 6y+a = 0,
\ncentre C = (2, -3) and radius = 4 (given)
\n\u2234 \\(\\sqrt{4+9-\\mathrm{a}}\\) = 4 \u21d2 13 – a = 16 \u21d2 a – 3.<\/p>\n

Question 8.
\nFind the equation of the circle whose extremities of diameter are (1, 2) and (4, 5).
\nSolution:
\nTaking A(1, 2) (x1<\/sub>,y1<\/sub>) and B(4, 5) (x2<\/sub>,y2<\/sub>) the equation of circle having A, B as extremities of diameter is
\n(x-x1<\/sub>)(x-x2<\/sub>)+(y-y1<\/sub>)(y-y2<\/sub>)=0
\n(x – 1)(x-4) + (y-2)(y-5) =0
\n= x2<\/sup>-5x+4+y2<\/sup>-7y+ 10 =0
\n= x2<\/sup>+y2<\/sup>-5x-7y+ 14=0<\/p>\n

\"TS<\/p>\n

Question 9.
\nFind the other end of the diameter of the circle x2<\/sup> + y2<\/sup>-8x – 8y + 27 = 0 if one end of it is\u00a0 (2, 3).
\nSolution:
\nCentre of the given circle is C (4, 4).
\nOne end of diameter is A = (2, 3). Let the other end be B (x, y). Then C is the end point of AB.
\n\u2234 \\(\\frac{x+2}{2}=4 \\) and \\( \\frac{y+3}{2}=4\\)
\n\u21d2 x = 6, y = 5
\n\u2234 Other end of the diameter B = (6, 5)<\/p>\n

Question 10.
\nObtain the parametric equations of x2<\/sup> + y2<\/sup> = 1.
\nSolution:
\nCentre of the circle = (0, 0) and radius = 1 = (h, k)
\nThe parametric equations of curve are
\nx = h + rcos\u03b8 = 0 + 1. cos\u03b8 = cos\u03b8
\ny = k + rsin\u03b8 = \u03b8+1.sin\u03b8 = sin\u03b8
\n0 \u2264 \u03b8 \u2264 2\u03c0<\/p>\n

Question 11.
\nObtain the parametric equation of the circle represented by x2<\/sup> + y2 <\/sup>+ 6x + 8y – 96 = 0.
\nSolution:
\nCentre(h,k) =(-3,-4)
\nand radius r = \\(\\sqrt{9+16+96}=\\sqrt{121}\\) = 1
\n\u2234 x = h + r cos\u03b8 = – 3 + 11 cos\u03b8
\ny = k + r sin 0=- 4 + 11 sin \u03b8,0\u22640\u2264 2\u03c0<\/p>\n

Question 12.
\nLocate the position of the point (2, 4) w.r.t circle x2<\/sup> + y2<\/sup> – 4x – 6y + 11 = 0.
\nSolution:
\nHere (x, y) (2, 4) and
\nx2<\/sup> + y2<\/sup> – 4x – 6y+ 11=0
\nS \u2261 (2)2<\/sup>+(4)2<\/sup>– 8 – 12 +11=-1
\nSince S1<\/sub>1<\/sub> <0, the point (2, 4) lies Inside the circle.<\/p>\n

\"TS<\/p>\n

Question 13.
\nFind the length of the tangent from (1,3) to the circle x2<\/sup> + y2<\/sup> -2x + 4y – 11 = 0.
\nSolution:
\nGiven (x1<\/sub>, y1<\/sub>) = (1, 3) and
\nS \u2261 x2<\/sup>+y2<\/sup>-2x+4y-11 = 0
\nS1<\/sub>1<\/sub> = 12<\/sup>+32 <\/sup>-2 + 12 – 11 = 9
\n\u2234 Length of the tangent from P(x1<\/sub>, y1<\/sub>) to S = 0 is
\n= \\(\\sqrt{S_{11}}=\\sqrt{9}\\) = 3<\/p>\n

Question 14.
\nShow that the circle S \u2261 x2<\/sup> + y2<\/sup>+2gx+2fy+ c = 0 touches (i) x-axis if g2<\/sup> = c (ii) Y-axis if f2<\/sup> =c.
\nSolution:
\n(i) We have the Intercept made by S = 0 on X-axis is \\(2 \\sqrt{\\mathrm{g}^2-\\mathrm{c}}\\).
\nIf the circle touches X-axis then \\(2 \\sqrt{\\mathrm{g}^2-\\mathrm{c}}\\) = \u21d2 g2<\/sup>= c.<\/p>\n

(ii) Similarly if the intercept made by S = 0 on
\nY-axis is \\(2 \\sqrt{f^2-c}\\) . If the circle touches Y-axis then \\(2 \\sqrt{f^2-c}\\) = f2<\/sup>=c.<\/p>\n

Question 15.
\nFind the equation of tangent to x2<\/sup> + y2<\/sup> – 6x +4y – 12 = 0 at (-1,1).
\nSolution:
\nWe have the equation of tangent at (x1<\/sub>, y1<\/sub>) to
\nS = 0 is xx1<\/sub> +yy1<\/sub> +g(x+x1<\/sub>)+1(y+y1<\/sub>)+ c
\n\u21d2 x(-1) 4y(1) – 3(x-1) + 2(y+ 1)-12 = 0
\n\u21d2 – 4x+3y-7 = 0 ,
\n\u21d2 4x – 3y+ 7 = 0<\/p>\n

Question 16.
\nShow that the line 5x + 12y – 4 = 0 touches the circle x2<\/sup> + y2<\/sup> – 6x + 4y + 12 = 0.
\nSolution:
\nCentre of the given circle = (3, -2) and
\nradius = \\(\\sqrt{9+4-12}=1\\)
\nThe perpendicular distance from the centre
\n(3,-2) to the line 5x + 12y- 4 = 0 is
\n\\(=\\left|\\frac{5(3)+12(-2)-4}{\\sqrt{25+144}}\\right|=\\left|\\frac{-13}{13}\\right|=1\\)
\n\u2234 radius of the circle.
\n\u21d2 The line 5x + 12y-.4 = 0 touches the given circle.<\/p>\n

Question 17.
\nFind the area of the triangle formed by the tangent at P(x1<\/sub>, y1<\/sub>)to the circle x2<\/sup> + y2<\/sup> = a2<\/sup> with the coordinate axes where x1<\/sub> y1<\/sub> \u2260 0.
\nSolution:
\nEquation of tangent at (x1<\/sub>, y1<\/sub>) to the circle
\nx2<\/sup> +y2<\/sup>-a2 <\/sup>is xx1<\/sub> +yy1<\/sub>– a2<\/sup>=0.
\nx, y intercepts are \\(\\frac{a^2}{x_1}\\) and \\(\\frac{\\mathrm{a}^2}{\\mathrm{y}_1}\\)
\n\u2234 Required area of the triangle
\n=\\(\\frac{1}{2}\\left|\\frac{a^2}{x_1} \\cdot \\frac{a^2}{y_1}\\right|=\\frac{a^4}{2\\left|x_1 y_1\\right|}\\)<\/p>\n

\"TS<\/p>\n

Question 18.
\nState the necessary and sufficient condition forlx+ my + 0 to be normal to the circle x2<\/sup>+ y2<\/sup> + 2gx + 2fy + c = 0.
\nSolution:
\nThe straight line lx + my + n = 0 is a normal to the circle S \u2261 x2<\/sup> + y2<\/sup> + 2 + 2fy + c = 0.
\n\u21d4 Centre (-g, – f) of the circle lies on lx + my + n = 0
\n\u21d4 l(-g) + m(-f) + n = 0
\n\u21d4 lg + mf = n<\/p>\n

Question 19.
\nFind the condition that the tangents are drawn from the exterior point (g,f) to S \u2261 x2<\/sup>+y2<\/sup>+ 2gx + 2fy + c= 0 are perpendicular to each other.
\nSolution:
\nIf the angle between the tangents drawn from P(x1<\/sub>,y1<\/sub>) to S=0 is \u03b8 then
\n\"TS<\/p>\n

Question 20.
\nFind the chord of contact of (2, 5) with respect to the circle x2<\/sup> + y2<\/sup> – 5x + 4y -2 = 0.
\nSolution:
\n2g=-5 and 2f = 4 \u21d2 g\\(\\frac{5}{2}\\) and f=2,c=-2
\nEquation of chord of contact of (x1<\/sub>, y1<\/sub>) w.r.t S = 0 is
\nxx1<\/sub> +yy1<\/sub> + g(x+x,) +f(y+y1<\/sub>)+c=0
\n=2x+5y-(x+2)+2(y+5)-2=0
\n= x-14y+6=0<\/p>\n

Question 21.
\nFind the equation of the polar of the point (2, a)w.r.tthe circle x2<\/sup>+y2<\/sup>+6x +8y-96 =0.
\nSolution:
\nEquation of polar of (x1<\/sub>, y1<\/sub>) (2, 3) is +yy +g(x+x1<\/sub>)+f(y+y1<\/sub>)+c=0
\n\u21d2 x(2) +y(3)+ 3(x + 2) +.4(y+ 3)- 96 = 0
\n\u21d2 5x + 7y – 78 = 0
\n\u21d2 (x1<\/sub>– a)2<\/sup>= (x1<\/sub> +a)2<\/sup>+y2<\/sup>
\n\u21d2 (x1<\/sub> – a)2<\/sup> – (x1<\/sub> + a)2<\/sup> y2<\/sup>1<\/sub>
\n\u21d2 y – 4ax1<\/sub> \u21d2\u00a0y2<\/sup>1<\/sub> + 4ax1<\/sub>\u00a0= 0
\n\u2234 Locus of (x1<\/sub>, y1<\/sub>) is y2<\/sup> + 4ax = 0.<\/p>\n

\"TS<\/p>\n

Question 22.
\nFind the pole of the line x+y+2 = 0 w.r.t x2<\/sup> + y2<\/sup>– 4x + 6y – 12=0.
\nSolution:
\nHere lx +my+n=0 is x+y+2=0 and S=0 is x2<\/sup> + y2<\/sup> – 4x + 6y – 12 = 0.
\n\"TS<\/p>\n

Question 23.
\nShow that (4, -2) and (3, -6) are conjugate w.r.t. the circle x2<\/sup> + y2<\/sup> – 24 = 0.
\nSolution:
\nHere (x1<\/sub>, y1<\/sub>) = (4, – 2) and (x2<\/sub>, y2<\/sub>) (3, -6) and S = x2<\/sup> + y2<\/sup>-24 = 0 ……….. (1)
\nTwo points (x1<\/sub>, y1<\/sub>) and (x2<\/sub>, y2<\/sub>) are conjugate w.r.t S=0 if S12<\/sub>=0
\n\u2234 x1<\/sub>x2<\/sub>+y1<\/sub>y1<\/sub>-24 = 0
\nFor the given points
\nS12 <\/sub>=4(3)+(-2)(-6)-24-0
\n\u2234 The given points are conjugate w.r.t the given circle.<\/p>\n

Question 24.
\nIf (4, k) and (2,3) are conjugate points w.r.t x2<\/sup> + y2<\/sup> = 17 then find k.
\nSolution:
\n(x1<\/sub>, y1<\/sub>)= (4, k) and (x2<\/sub>, y2<\/sub>)= (2, 3). Since the given points are conjugate S12 <\/sub>= 0.
\n= x1<\/sub>x2<\/sub> + y1<\/sub>y2<\/sub> – 17 = 0
\n(4)(2)+(k)(3)-17=0 \u21d2 k=3<\/p>\n

Question 25.
\nShow that the lines 2x+3y+ 11 =0,and 2x – 2y -1= 0 are conjugate w.r.t x2<\/sup> + y2<\/sup>+ 4x + 6y + 12 = 0.
\nSolution:
\n\"TS
\nQuestion 26.
\nFind the inverse point of (2, -3) wrt the circle x2<\/sup>+y2<\/sup>– 4x-6y+9=0
\nSolution:
\nLet P(-2, 3) and C (2,3) is the centre of the given circle. Then the polar of P is
\nx(-2)+y(3)-2(x-2)-3(y+3)+9=0
\nx = 1 ……………. (1)
\nEquation of line \\(\\overline{\\mathrm{CP}}=\\mathrm{y}-3=\\frac{3-3}{2+1}(\\mathrm{x}+2)\\)
\n\u21d2 y – 3 = 0 \u21d2 y = 3 ……………. (2)
\n\u2234 From (1) and (2) the inverse point of
\nP(-2, 3) is (1, 3).<\/p>\n

\"TS<\/p>\n

Short Answer Type Questions<\/span><\/p>\n

Question 1.
\nFrom the point A(0, 3) on the circle x2<\/sup> + 4x +(y-3)2<\/sup>= 0 a chord AB is drawn and extended to a point M such that
\nAM = 2AB. Find the equation to the locus of M.
\nSolution:
\n\"TS<\/p>\n

Let M (x1<\/sub>, y1<\/sub>) be the locus. Given AM – 2A8
\n= AB+ BM = AB+AB
\nBM – AB \u21d2 B is the mid point of AM
\n\"TS<\/p>\n

Question 2.
\nFind the equation of the circle passing through (4, 1), (6,5) and having the centre on the line 4x+y-16=0.
\nSolution:
\nLet the equation of the required circle be
\nx2<\/sup>+y2<\/sup>+2gx+2fy+c = 0 ……………. (1)
\nSince it passes through (4, 1) we have
\n16 + 1 + 8g + 2f + c = 0
\n= 17+8g+2f+c=0 ……………. (2)
\nSimilarly (6, 5) lies on (1) then
\n36+25+12g+ 10f +c=0
\n= 61+12g+10f+c=0 ……………. (3)
\nGiven that the centre of circle (-g, -f) lies on 4x + y-16 = 0
\n-4g-f-16=0
\n\u21d2 4g+f+16=0 ……………. (4)
\nFrom (2) and (3)
\n– 44 – 4g – 8f 0
\n=g+2f=- 11\u00a0 ………….. (5)
\nFrom (4) 4g+ f =- 16
\n\"TS
\n\u21d2 g= – 3, f =- 4 and
\nfrom (2)
\n17 – 24 – 8+c – 0 \u21d2 c = 15
\n\u2234 Equation of the required circle from (1) is
\nx2<\/sup> +y2<\/sup> – 6x – 8y + 15=0.<\/p>\n

\"TS<\/p>\n

Question 3.
\nSuppose a point (x1<\/sub>, y1<\/sub>) satisfies x2<\/sup> + y2<\/sup> + 2gx + 2fy + c= 0 then show that it represents a circle whenever g, f and c are real.
\nSolution:
\nComparing the given equation with ax2<\/sup> + 2hxy + by2<\/sup> + 2gx + 2fy + c = 0, we have coefficient of x2<\/sup> = coefficient of y2<\/sup> and coefficient of xy term = 0.
\nThe given equation represents a circle If g2<\/sup> + f2<\/sup> \u2265 0
\nSince (x1<\/sub>, y1<\/sub>) is a point on the circle we have
\nx1<\/sub> + y1<\/sub> + 2gx1<\/sub> + 2fy1<\/sub> + c = 0
\ng2<\/sup> + f2 <\/sup>– c = g2<\/sup> + f2<\/sup> +x2<\/sup>1<\/sub> +y2<\/sup>1<\/sub> + 2gx1<\/sub> + 2fy1 <\/sub>= (x1<\/sub> +g) + (y1<\/sub> + f)2<\/sup> \u2265 0
\nSince g, f and c are real the equation (1) represents a circle.<\/p>\n

Question 4.
\nFind the equation of circle which Louches x-axis at a distance of 3 from the origin and making intercept of length 6 on y-axis.
\nSolution:
\n\"TS
\nLet the equation of required circle be
\nx2<\/sup>+y2<\/sup>+2gx+2fy+c=0 …………………….. (1)
\nIf it touches x- axis at (3, 0) then 9 + 0 + 6g + c = 0
\n\u21d2 6g+c= – 9 …………………… (2)
\nIf circle touches x-axis then g2 <\/sup>– c = 0 ………………………. (3)
\nAdding (2) and (3)
\ng2<\/sup> + 6g = -9
\n= (g+3)2 <\/sup>= 0 = g = -3 ……………………. (4)
\n\u2234 From (3), C = 9
\nAlso given that intercept on y-axis is 6
\n\"TS<\/p>\n

Question 5.
\nFind the equation of circle which passes through the vertices of the triangle formed by
\nL1<\/sub> =x+y+ 1 =0, L2<\/sub>=3x+y-5=0 and L3<\/sub> = 2x + y-5 = 0.
\nSolution:
\nSuppose L1<\/sub>, L2<\/sub>; L2<\/sub>, L.3<\/sub> and L3<\/sub>, L1<\/sub> intersect at A, B and C respectively. Consider a curve whose equation is
\nk(x+y+1)(3x+y-5)+1(3x+y-5)
\n(2x+y-5)+m(x+y+ 1)
\n(2x+y-5) = 0 ……………………. (1)
\nWe can verify that this curve passes through A, B, C. So we find k, I and m such that the equation (1) represents a circle. If (1) represents a circle then
\n(i) coefficient of x2<\/sup> = coefficient of y2<\/sup>
\n= 3k + 6l + 2m = k + l + m
\n= 2k+5l+m=0 ………………… (2)
\n(ii) coefficient of xy is zero.
\n4k+5l+3m= 0 ………………… (3)
\nSolving (2) and (3) we get
\n\"TS
\nHence the required equation is
\n5(x+y+1)(3x+y-5)-1(3x+y-5)
\n(2x+y-5)-5(x+y+1)
\n(2x + y -5) = 0
\n\u21d2 x2<\/sup>+y2<\/sup>-30x-10y+ 25=0<\/p>\n

\"TS<\/p>\n

Question 6.
\nFind the centre of the drive passing through the points (0,0), (2,0) and (0,2).
\nSolution:
\nLet the equation of required circle be
\nx2<\/sup>+y2<\/sup>+2gx+2fy+c=0 ……………….. (1)
\nIf (1) passes through (0, 0) then c = 0
\nIf (1) passes through (2, 0) then
\n4+4g+c=0 ………….. (1)
\nIf (1) passes through (0, 2) then
\n4+4f+c=0 ………….. (2)
\nFrom (2) and (3) we have
\ng=-1 and f = – 1 (\u2235 c=0)
\n\u2234 Centre of the circle = (-g, -f) (1, 1)<\/p>\n

Question 7.
\nIf a point P is moving such that the length of tangents drawn from P to x2<\/sup>+y2<\/sup>\u00a0– 2x + 4y – 20 = 0
\nx2<\/sup>+y2<\/sup> – 2x-8y+ 1=0 are in the ratio 2: 1 then show that the equation of the locus of P is x2<\/sup>+y2 <\/sup>-2x – 12y+8=0.
\nSolution:
\nLet P (x1<\/sub>, y1<\/sub>) be the locus and \\(\\overline{\\mathrm{PT}_1}, \\overline{\\mathrm{PT}_2}\\) are the tangents drawn from the points P to the two circles x2<\/sup>+y2<\/sup>\u00a0– 2x + 4y – 20 = 0 and x2<\/sup>+y2<\/sup>– 2x – 8y + 1 = 0
\n\"TS<\/p>\n

Question 8.
\nlf S\u2261 x2<\/sup>+y2<\/sup>+2gx+2fy+c=0 represents a circle then show that the straight line lx + my + n = 0.
\n(i) touches the circle S = 0 if
\n\\(\\left(g^2+f^2-c\\right)=\\frac{(g l+m f-n)^2}{\\left.l^2+m^2\\right)}\\)<\/p>\n

(ii) meets the circle S=0 in two points if.
\n\\(g^2+f^2-c>\\frac{(g l+m f-n)^2}{\\left.a^2+m^2\\right)}\\)<\/p>\n

\"TS<\/p>\n

(ii) will not meet the circle if
\n\\(\\mathrm{g}^2+\\mathrm{f}^2-\\mathrm{c}<\\frac{(\\mathrm{g} l+\\mathrm{mf}-\\mathrm{n})^2}{\\left.l^2+\\mathrm{m}^2\\right)}\\)
\nSolution:
\n(i) The given straight line lx + my + n = 0
\ntouches the circle S \u2261 x2<\/sup>+y2<\/sup> + 2gx + 2fy + c = 0 if the perpendicular distance from (-g, -f) to lx + my + n – 0 is equal to radius r.
\n\"TS<\/p>\n

(ii) The given line meets the circle S=0 in two points
\n\\(\\mathrm{g}^2+\\mathrm{f}^2-\\mathrm{c}>\\frac{(\\mathrm{lg}+\\mathrm{mf}-\\mathrm{n})^2}{l^2+\\mathrm{m}^2}\\)<\/p>\n

(iii) The given line will not meet the circle S=0
\nIf \\(\\mathrm{g}^2+\\mathrm{f}^2-\\mathrm{c}<\\frac{(\\mathrm{gl} l \\mathrm{mf}-\\mathrm{n})^2}{l^2+\\mathrm{m}^2}\\)<\/p>\n

Question 9.
\nFind the length of the chord intercepted by the circle x2<\/sup>+y2<\/sup>+8x-4y – 16 = 0 on the line 3x-y+4 = 0.
\nSolution:
\n\"TS
\nCentre of the circle C = (-4, 2) and
\nradius= \\(\\sqrt{16+4+16}=6\\)
\nCL = Perpendicular distance from C(-4,2) to the chord 3x-y + 4=0.
\n\"TS<\/p>\n

Question 10.
\nFind the equation of tangents to x2<\/sup>+y2<\/sup>-4x+6y-12=0 which are parallel to x + 2y -8 = 0.
\nSolution:
\nCentre of the given circle C = (2, -3) and radius \\(\\sqrt{4+9+12}\\) =5
\nAny line parallel to x + 2y – 8= 0 is of the form x + 2y + k – 0. If this line becomes a tangent then the perpendicular distance from C(2, -3) to x + 2y + k = 0 is equal to the radius.
\n\u2234 \\(\\left|\\frac{2-6+k}{\\sqrt{1+4}}\\right|=5\\)
\n\u21d2 |k – 4| = 5\\(\\sqrt{5}\\) \u21d2 k = 4 \u00b1 5\\(\\sqrt{5}\\)
\n\u2234 Equation of parallel tangents are
\nx+2y+(4\u00b15\\(\\sqrt{5}\\) \u00a0)=0<\/p>\n

Question 11.
\nFind the equation of tangent to x2<\/sup>+y2 <\/sup>– 2x+ 4y = 0 at (3, -1). Also find the equation of tangent parallel to it.
\nSolution:
\nEquation of tangent at (3, -1) to the circle
\nx2<\/sup>+y2 <\/sup>-3x + = 0 is
\nx(3)+y(-1)-1(x+3)+2(y-1)=0
\n= 3x-y-1(x+ 3) + 2(y-1) 0
\n3x-y-x-3+2y-20
\n2x+y-5=0 ……………………. (1)
\nEquational line parallel to 2x + y-5 = 0 is of the from 2x + y + k = 0. If this is a tangent to the given circle then the perpendicular distance from the centre (1, -2) is equal to the radius=\\(\\sqrt{1+4}=\\sqrt{5}\\)
\n\u2234 \\(\\left|\\frac{2(1)-2+k}{\\sqrt{5}}\\right|=\\sqrt{5}\\)
\n= k -\u00b15
\n\u2234 Equations of parallel tangents to (1) are 2x + y \u00b1 5 = 0
\n\u2234 The equation of other parallel tangent is 2x + y + 5 = 0<\/p>\n

Question 12.
\nIf 4x-3y+7=0 is a tangent to the circle represented by x2<\/sup>+y2<\/sup>-6x+4y-12=0 then find the point of contact.
\nSolution:
\nCentre of the given circle C = (3, -2). Let the P(x1<\/sub>, y1<\/sub>) be the contact.
\nThen 4x1<\/sub> – 3y1<\/sub> + 7 = 0 …………… (1) is perpendicular to PC, equation of PC is 3x + 4y + k = 0
\nSince this passes through C(3, -2) we have
\n9 – 8+k=0=k=-1
\n\u2234 Equation of CP is 3x1<\/sub> + 4y1<\/sub> – 1 = 0 ……………. (2)
\nSolving (1) and (2)
\n\"TS
\n\u2234 P(-1, 1) is the point of contact.<\/p>\n

\"TS<\/p>\n

Question 13.
\nFind the equations of circles which touch 2x – 3y+ 1 =0 at (1, 1) and having radius \\(\\sqrt{13}\\)
\nSolution:
\n\"TS
\nEquation of line \u22a5r to 2x – 3y 1 – 0 is of the from 3x + + k -0; Since this passes through
\n(1, 1) we have 3+2 +k=0 \u21d2 k=-5
\nEquation of line perpendicular to the tangent is 3x+2y-5=0 ………….. (1)
\nLet (x, y) be the centre of circle
\n\"TS
\n\u21d2 x2 <\/sup>– 2x-3=0 \u21d2 (x-3)(x+1)=0
\n\u21d2 x = 3 or x = – 1
\nWhen x = 3, we have from (1) y =\\(\\frac{5-9}{2}\\) – 2
\nand when x=-1, y = \\(\\frac{5+3}{2}=4\\)
\n\u2234 Centre are (3, -2) and (-1, 4).
\n\u2234 Equations of circles with (3, – 2) and
\n(-1, 4) With radius ,\\(\\sqrt{13}\\) are given by
\n(x-3)2<\/sup> + (y+ 2)2 <\/sup>= 13 and (x+ 1)2<\/sup> + (3,4)2<\/sup> = 13
\n= x2<\/sup> +y2<\/sup>-6x+4y=0 and x2<\/sup> +y2<\/sup>-2x-8y+4 = 0.<\/p>\n

\"TS<\/p>\n

Question 14.
\nFind the equation of the tangent at the point 300 (Parametric value of O) of the circle x2 4-y2+4x+6y-39=0.
\nSolution:
\nHere g=2, f=3. c=-39
\n\\(r=\\sqrt{4+9+39}=\\sqrt{52}=2 \\sqrt{13}\\)
\nThe required equation of tangent at \u2018\u03b8\u2019 to
\nS = 0 is given by the formula
\n\"TS<\/p>\n

Question 15.
\nFind the equation of normal to the circle x2<\/sup>+y2<\/sup>– 4x-6y+ 11 =0 at (3, 2), Also find the other point where the normal meets the circle.
\nSolution:
\nLet A(3, 2) and C be the centre of the given circle C = (2, 3) = (-g,-f)
\nEquation of normal at (x1<\/sub>, y1<\/sub>) is (x-x1<\/sub>)(y1<\/sub>+f)-(y-y1<\/sub>)(x1<\/sub>+g) = 0
\n\u21d2 (x-3)(2-3)-(y-2)(3-2)=0
\n\u21d2 1(x-3)- 1(y-2)= 0
\n\u21d2 x-y+5 =0 x+y-5 =0
\nLet B (x1<\/sub>, y1<\/sub>) be the other point where the normal meets the circle.
\nThen \\(\\frac{x_1+3}{2}=2\\) and \\(\\frac{y_1+2}{2}=3\\)
\nx1<\/sub> = 1, and y1<\/sub> = 4
\nHence normal at (3,2) meets the circle at (1,4).<\/p>\n

Question 16.
\nFind the area of the triangle formed by the normal at (3, -4) to the circle x2<\/sup>+y2<\/sup>-22x-4y+ 25=0 with the coordinate axes.
\nSolution:
\nFrom the given equation of circle
\n2g = -22 and 2f = – 4=g=-11 and f=-2
\nAlso (x1<\/sub>, y1<\/sub>) =(3,-4)
\nThen equation of normal al (x1<\/sub>, y1<\/sub>) is
\n(x-x1<\/sub>)(y-y1<\/sub>)-(y-y1<\/sub>)(x1<\/sub> +g)=0
\n\u21d2 (x-3)(-4-2)-(y+4)(3-11)=0
\n\u21d2 (x-3)(-6)-(y+4)(-8)=0
\n\u21d2 -6x + + 50 = 0
\n\u21d2 3x – 4y – 25 = 0
\n\"TS<\/p>\n

Question 17.
\nIf \u03b81<\/sub>, \u03b82<\/sub> are the angles of inclination of tangents through a point P to (lie circle x2<\/sup> + y2<\/sup>= a2<\/sup>\u00a0then find the locus of P where cot \u03b81<\/sub>+ cot \u03b82<\/sub> = k
\nSolution:
\nThe equation of tangent to the circle x2<\/sup> + y2 <\/sup>= a2<\/sup> having slope m is y = mx + \\(a \\sqrt{1+m^2}\\)
\nLet P(x1<\/sub>, y1<\/sub>) be a point on the locus. Then
\n\"TS<\/p>\n

Question 18.
\nIf the chord of contact of P with respect to the circle x2<\/sup> + y2<\/sup> = a2<\/sup> cut the circle at A and B such that \\(\\angle \\mathrm{AOB}=90^{\\circ}\\) then show that P lies on the circle x2<\/sup>+y2<\/sup>=2a2<\/sup>.
\nSolution:
\n\"TS
\nLet P(x1<\/sub>, y1<\/sub>) be a point and let the chord of contact of P(x1<\/sub>, y1<\/sub>) meets circle are A and B.Such that \\(\\angle \\mathrm{AOB}=90^{\\circ}\\)
\nEquation of chord of contact of P(x1<\/sub>, y1<\/sub>) is
\nxx1<\/sub> + yy1<\/sub> – a2<\/sub> = 0 \u21d2 \\(\\frac{x_1+y y_1}{a^2}=1\\) ……….. (1)
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 19.
\nShow that the poles of tangents to the circle x+y = a2<\/sup> w.r.t. the circle (x + a)2<\/sup> + 2 = 11 on y2<\/sup> + 4ax = 0.
\nSolution:
\nLet P(x1<\/sub>, y1<\/sub>) be the pole of the tangent to the circle x2<\/sup> + y2<\/sup> – a2<\/sup> …………………… (1) w.r.t circle (x + a)2<\/sup> + y2<\/sup> = 2a2<\/sup>. Then the equation of polar of P (x1<\/sub>, y1<\/sub>) w.r.t
\n(x + a)2<\/sup> + y2<\/sup> . 2a2<\/sup> is …………………. (2)
\nxx1<\/sub> +yy1<\/sub> +a(x1<\/sub>+ y1<\/sub>)-a2<\/sup>= 0
\n= x(x1<\/sub> + a) + yy1<\/sub> + (ax1<\/sub> – a2<\/sub>) = 0 ………………. (3)
\nThe line is a tangent to the circle (1) then perpendicular distance from (0, 0) to (3) is equal to radius \u2018a\u2019.
\n\"TS<\/p>\n

Question 20.
\nShow that the area of the triangle formed by the two tangents through P(x1<\/sub>, y1<\/sub>) to the circle S=x2<\/sup>+y2<\/sup>+2gx+2fy+c=0 and the chord of contact of P.w.r. IS=0 is \\(\\frac{r\\left(S_{11}\\right)^{\\frac{3}{2}}}{S_{11}+r^2}\\)\u00a0 where r is the radius of the circle.
\nSolution:
\n\"TS<\/p>\n

Let PA and PB be the two tangents drawn from P(x), y) to the circle S = 0 and \u03b8 be the angle between these two tangents.
\nThen tan = \\(\\frac{\\theta}{2}=\\frac{r}{\\sqrt{S_{11}}}\\)
\nArea of triangle formed by the tangents through P(x1<\/sub>, y1<\/sub>) to S = 0 and the chord of contact of P w.r.t S = 0
\n\"TS
\nQuestion 21.
\nFind the mid point of the chord Intercepted by x2<\/sup>+ y2<\/sup> – 2x – 10y + 1 = 0 on the line x – 2y + 7 = 0.
\nSolution:
\nLet x2<\/sup>+ y2<\/sup> – 2x – 10y + 1 = 0 ………………. (1)
\nx – 2y+ 7= 0 ……………… (2)
\nLet P(x1<\/sub>, y1<\/sub>) be the midpoint of the chord intercepted by the circle (L) on the line given by (2).
\nThe equation of chord of (x1<\/sub>, y1<\/sub>) in terms of its midpoint is
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 22.
\nFind the equation of pair of tangents drawn from (10, 4) to the circle x2<\/sup> + y2<\/sup> = 25.
\nSolution:
\nEquation of pair of tangents is S. S1<\/sub>1 <\/sub>= S1<\/sub>2<\/sup>
\n\u21d2 (100+16-25)(x2<\/sup>+y2<\/sup>-25)=(10x+4y-25)2<\/sup>
\n\u21d2 91(x2<\/sup>+ y2<\/sup>-25)=100x2<\/sup>+16y2<\/sup>+625+80xy – 200y – 500x
\n\u21d2 9x2<\/sup> +80xy-75y2<\/sup>-500x-200y + 2900=0<\/p>\n

Long Answer Type Questions<\/span><\/p>\n

Question 1.
\nFind the equation of circle passing through P(1, 1), Q(2, -1) and R(3, 2).
\nSolution:
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 2.
\nFind the equation of the circumcircle of the triangle formed by the line ax + by + c=0 (abc \u2260 0) and the coordinate axes.
\nSolution:
\nLet the line ax+by+c=0 cuts x, and y axis at A and B so that
\n\"TS
\nare the vertices of the triangle.
\nLet x2<\/sup>+y2<\/sup>+2gx+2fy+c=0 …………… (1) be the required equation of the circle. Since it passes through (0, 0) we have c= 0.
\n\"TS<\/p>\n

Question 3.
\nFind the locus of mid points of the chords of contant of x2<\/sup> + y2<\/sup> = a2<\/sup> from the points lying on the line lx + my + n=0.
\nSolution:
\nLet (x1<\/sub>, y1<\/sub>) be the locus of mid points of chords of the circle x2<\/sup> + y2<\/sup> = a2<\/sup> ………………. (1)
\nand this is a chord lies on ix + my + n = 0 ………………… (2)
\ni.e., pole of this chord is on (2).
\nEquation of chord of (1) having (x1<\/sub>, y1<\/sub>) as its mid point is xx1<\/sub> + yy1<\/sub> – x + y
\n\"TS<\/p>\n

Question 4.
\nShow that four common tangents can be drawn for the circles given by
\nx2<\/sup>+y2<\/sup>-14x+6y+33=9 ……………… (1)
\nx2<\/sup>+y2<\/sup>+30x-2y+1=0………………(2)
\nand find the Internal and external centres of similitude.
\nSolution:
\nCentre of circLe (1) is C1<\/sub> = (7, – 3)
\nCentre of circle (2) is C2<\/sub> = (15, 1)
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 5.
\nProve that the circles
\nx2<\/sup>+y2<\/sup>-8x-6y+21=0 ……………. (1) and x2<\/sup>+y2<\/sup>-2y-15=0 ………………… (2) have exactly two common tangents.
\nAlso find the point of Intersection of those tangents.
\nSolution:
\nCentres of circles are C1<\/sub> (4, 3) and C2<\/sub> – (0, -1)
\n\"TS
\n\u2234 Given circles intersect each other and have exactly two common tangents.
\nr1<\/sub> : r2<\/sub> = 2 : 4 = 1: 2
\nThe point of contact P divides C1<\/sub>C2<\/sub> externally\u00a0in the ratio 1: 2.
\n\u2234 External centre of similitude
\n= \\(\\left(\\frac{8-0}{2-1}, \\frac{6-1}{2-1}\\right)=(8,5)\\)<\/p>\n

Question 6.
\nShow that the circles
\nx2<\/sup>+y2<\/sup>– 4x-6y-12=0 ………………. (1)
\nand x2<\/sup>+y2<\/sup>+6x+18y+26=0 …………….. (2) touch each other. Also find the point of contact and common Langent at this point of contact.
\nSolution:
\nLet C1<\/sub> (2, 3) and C2<\/sub> (- 3, -9) are centres of circles (1) and (2) and their radii are
\n\"TS
\nC1<\/sub>C2<\/sub> = r1<\/sub> + r2<\/sub> and hence the two circles touch each other externally. Point of contact divides C1<\/sub>C2<\/sub> in the ratio
\nr1<\/sub> : r2<\/sub> = 5 : 8
\n\"TS<\/p>\n

Question 7.
\nShow that the circles x2<\/sup>+y2<\/sup>-4x-6y-12=0 and 5(x2<\/sup>+y2<\/sup>)-8x-14y-32=0 touch each other and find their point of contact.
\nSolution:
\n\"TS
\nNow \\(\\overline{\\mathrm{C}_1 \\mathrm{C}_2}\\)\u00a0= | r1<\/sub>-r2 <\/sub>|
\nHere the circles (1) and (2) touch each other internally the point of contact P dividies C1<\/sub>C2<\/sub> in the ratio 5 : 3 externally.
\n\\(\\mathrm{P}=\\left(\\frac{3(2)-5\\left(\\frac{4}{5}\\right)}{3-5}, \\frac{3(3)-5\\left(\\frac{7}{5}\\right)}{3-5}\\right)\\)
\nE (-1, – 1)
\n\u2234 Point of contact = (-1, -1)<\/p>\n

\"TS<\/p>\n

Question 8.
\nFind the equations to all possible common tangents of the circles
\nx2<\/sup>+y2 <\/sup>-2x-6y+6=0 ………………. (1) and x2<\/sup>+y2 <\/sup>= 1 …………………. (2)
\nSolution:
\nCentres of circles are C1<\/sub> – (1, 3) and C2<\/sub> (0, 0)
\n\"TS
\n= 4y + 3xy- 9y – 3x + 5 = 0
\n= (y +l) (4y + 3x + m) (Suppose)
\nEquating the coefficient of x, y and constant terms
\n3l = – 3 ………………. (3)
\nand 4l + m = – 9 ……………..(4)
\nlm=5 …………….. (5)
\nFrom (3) and (4) l=-1\u21d2m=-5
\nEquations of transverse common tangents are (y-1) = 0 and 4y+3x-5=0
\nDirect common tangents are given by
\n(x2<\/sup>+y2<\/sup>-1)(1+9-1)=(xi-3y+ 1)2<\/sup>
\n=9(x2<\/sup>+y2<\/sup>-1)=x2<\/sup>+9y2<\/sup>+1+6xy+6y+2x
\n8x2<\/sup>– 6xy-2x – 6y- 10=0
\n= (x+l) (8x-6y+m)
\nComparing coefficient of x, y and constant and
\n8l+m=-2
\nand -6l=-6 =l= l and
\nlm = – 10 \u21d2 m – 10
\nEquations of direct common tangents are
\nx+ 1-0 and 8x-6y-10-0
\n\u21d2 4x – 3y – 5 = 0<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these\u00a0TS Inter 2nd Year Maths 2B Important Questions Chapter 1 Circles to help strengthen their preparations for exams. TS Inter 2nd Year Maths 2B\u00a0Circles Important Questions Very Short Answer Type Questions\u00a0 \u00a0 \u00a0\u00a0 Question 1. Find the equation of circle with centre (1, 4) and radius 5. Solution: Standand equation of circle … Read more<\/a><\/p>\n","protected":false},"author":2,"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\/13141"}],"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\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=13141"}],"version-history":[{"count":20,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/13141\/revisions"}],"predecessor-version":[{"id":13870,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/13141\/revisions\/13870"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=13141"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=13141"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=13141"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}