TS Inter 2nd Year Maths 2B Important Questions<\/a> Chapter 1 Circles to help strengthen their preparations for exams.<\/p>\nTS Inter 2nd Year Maths 2B\u00a0Circles Important Questions<\/h2>\n
Very Short Answer Type Questions\u00a0 \u00a0 \u00a0\u00a0<\/span><\/p>\nQuestion 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<\/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>\nQuestion 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<\/p>\nQuestion 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<\/p>\n
<\/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>\nQuestion 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>\nQuestion 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>\nQuestion 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<\/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>\nQuestion 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>\nQuestion 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>\nQuestion 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<\/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>\nQuestion 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>\nQuestion 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>\nQuestion 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>\nQuestion 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<\/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>\nQuestion 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<\/p>\nQuestion 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>\nQuestion 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<\/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<\/p>\nQuestion 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>\nQuestion 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>\nQuestion 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
\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<\/p>\n
Short Answer Type Questions<\/span><\/p>\nQuestion 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<\/p>\nLet 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<\/p>\nQuestion 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
\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<\/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>\nQuestion 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
\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<\/p>\nQuestion 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 L