{"id":35514,"date":"2022-09-19T17:17:54","date_gmt":"2022-09-19T11:47:54","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35514"},"modified":"2022-09-19T17:17:54","modified_gmt":"2022-09-19T11:47:54","slug":"inter-2nd-year-maths-2b-system-of-circles-solutions-ex-2b","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-2nd-year-maths-2b-system-of-circles-solutions-ex-2b\/","title":{"rendered":"Inter 2nd Year Maths 2B System of Circles Solutions Ex 2(b)"},"content":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions<\/a> Inter 2nd Year Maths 2B System of Circles Solutions Exercise 2(b) will help students to clear their doubts quickly.<\/p>\n

Intermediate 2nd Year Maths 2B System of Circles Solutions Exercise 2(b)<\/h2>\n

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

Question 1.
\nFind the equation of the radical axis of the following circles,
\ni) x\u00b2 + y\u00b2 – 3x – 4y + 5 = 0, 3(x\u00b2 + y\u00b2) – 7x + 8y – 11 = 0
\nSolution:
\nS \u2261 x\u00b2 + y\u00b2 – 3x – 4y + 5 = 0
\nS \u2261 3x\u00b2 + 3y\u00b2 – 7x + 8y + 11 = 0
\nS – S’ = 0 is radical axis.
\n(x\u00b2 + y\u00b2 – 3x – 4y + 5)
\n\"Inter<\/p>\n

ii) x\u00b2 + y\u00b2 + 2x + 4y + 1 = 0, x\u00b2 + y\u00b2 + 4x + y = 0.
\nSolution:
\nS – S’ = 0 is radical axis.
\n(x\u00b2 + y\u00b2 + 2x + 4y + 1) – (x\u00b2 +y\u00b2 + 4x + y) = 0
\n\u21d2 -2x + 3y + 1 = 0
\n(or) 2x – 3y – 1 =0 required radical axis.<\/p>\n

iii) x\u00b2 + y\u00b2 +4x + 6y – 7 = 0,
\n4(x\u00b2 + y\u00b2) + 8x + 12y – 9 = 0.
\nSolution:
\nS – S’ = 0 is radical axis.
\n(x\u00b2 + y\u00b2 + 4x + 6y – 7) – (x\u00b2 + y\u00b2 + 2x + 3y – \\(\\frac{9}{4}\\)) = o
\n\u21d2 2x + 3y – \\(\\frac{19}{4}\\) = 0 \u21d2 8x + 12y – 19 = 0<\/p>\n

iv) x\u00b2 + y\u00b2 – 2x – 4y -1=0, x\u00b2 + y\u00b2 – 4x – 6y + 5 = 0.
\nSolution:
\nS – S’ = 0 radical axis
\n(x\u00b2 + y\u00b2 – 2x – 4y – 1) – (x\u00b2 + y\u00b2 – 4x – 6y + 5) = 0
\n2x + 2y – 6 = 0 (or) x + y – 3 = 0<\/p>\n

\"Inter<\/p>\n

Question 2.
\nFind the equation of the common chord of the following pair of circles.
\ni) x\u00b2 + y\u00b2 – 4x – 4y + 3 = 0, x\u00b2 + y\u00b2 – 5x – 6y + 4 = 0.
\nSolution:
\n(x\u00b2 + y\u00b2 – 4x – 4y + 3) – (x\u00b2 + y\u00b2 – 5x – 6y + 4) = 0
\nx + 2y – 1 = 0 Equation of common chord.<\/p>\n

ii) x\u00b2 + y\u00b2 + 2x + 3y + 1 = 0, x\u00b2 + y\u00b2 + 4x + 3y + 2 = 0.
\nSolution:
\n(x\u00b2 + y\u00b2 +2x + 3y + 1) – (x\u00b2 + y\u00b2 + 4x + 3y + 2) = 0
\n-2x – 1 = 0 equation of common chord is<\/p>\n

iii) (x – a)\u00b2 + (y – b)\u00b2 = c\u00b2, (x – b)\u00b2 + (y – a)\u00b2 = c\u00b2 (a \u2260 b)
\nSolution:
\n(x\u00b2 + y\u00b2 – 2xa – 2yb – c\u00b2) – (x\u00b2 + y\u00b2 – 2xb – 2ya – c\u00b2) = 0
\n-2x(a – b) – 2y(b – a) = 0
\n(or) x – y = 0<\/p>\n

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

Question 1.
\nFind the equation of the common tangent of the following.circles at their point of contact.
\ni) x\u00b2 + y\u00b2 + 10x – 2y + 22 = 0, x\u00b2 + y\u00b2 + 2x – 8y + 8 = 0.
\nSolution:
\nx\u00b2 + y\u00b2 + 10x – 2y + 22 = 0
\nx\u00b2 + y\u00b2 + 2x – 8y + 8 = 0
\nWhen circles touch each other then
\nS – S’ = 0 is required tangent (common)
\n\u2234 (x\u00b2 + y\u00b2 + 10x-2y + 22) – (x\u00b2 + y\u00b2 + 2x – 8y + 8) = 0
\n8x + 6y + 14 = 0 (or)
\n4x + 3y + 7 = 0<\/p>\n

ii) x\u00b2 + y\u00b2 – 8y – 4 = 0; x\u00b2 + y\u00b2 – 2x – 4y = 0.
\nSolution:
\nWhen circles touch each other then
\nS – S’ = 0 is required common tangent.
\n(x\u00b2 + y\u00b2 – 8y – 4) – (x\u00b2 + y\u00b2 – 2x – 4y) = 0
\n2x – 4y – 4 = 0 (or) x – 2y – 2 = 0<\/p>\n

Question 2.
\nShow that the circles x\u00b2 + y\u00b2 – 8x – 2y + 8 = 0 and x\u00b2 + y\u00b2 – 2x +, 6y + 6 = 0 touch each other and find the point of contact.
\nSolution:
\nC1<\/sub> = (4, 1) C2<\/sub> = (1, -3)
\nr1<\/sub> = \\(\\sqrt{16+1-8}\\) = 3 ; r2<\/sub> = \\(\\sqrt{1+9-6}\\) = 2
\nC1<\/sub>C2<\/sub> = \\(\\sqrt{(4-1)^{2}+(1+3^{2})}\\) = 5
\nr1<\/sub> + r2<\/sub> = C1<\/sub> + C2<\/sub> they touch each other externally
\n\"Inter
\n\u2234 Point of contact is (\\(\\frac{11}{5} , \\frac{-7}{5}\\))<\/p>\n

Question 3.
\nIf the two circles x\u00b2 + y\u00b2 + 2gx + 2fy = 0 and x\u00b2 + y\u00b2 + 2g’x + 2f’y = 0 touch each other then show that f’g = fg’.
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 4.
\nFind the radical centre of the following circles.
\ni) x\u00b2 + y\u00b2 – 4x – 6y + 5 = 0 ………… (i)
\nx\u00b2 + y\u00b2 – 2x – 4y – 1 = 0 ………… (ii)
\nx\u00b2 + y\u00b2 – 6x – 2y = 0 ………… (iii)
\nSolution:
\n(i) – (ii) gives
\n– 2x – 2y + 6 = 0
\nx + y – 3 = 0 ………. (1)
\n(ii) – (iii) gives
\n\"Inter
\nPoint of intersection of (1) and (2) is radical centre will (7\/6,11\/6) we get by solving these two equations.<\/p>\n

ii) x\u00b2 + y\u00b2 + 4x – 7 = 0,
\n2x\u00b2 + 2y\u00b2 + 3x + 5y – 9 = 0,
\nx\u00b2 + y\u00b2 + y = 0.
\nSolution:
\nS = x\u00b2 + y\u00b2 + 4x – 7 = 0 ……….. (i)
\nS1<\/sub> = 2x\u00b2 + 2y\u00b2 + 3x + 5y – 9 = 0
\n= x\u00b2 + y\u00b2 + \\(\\frac{3}{2}\\)x + \\(\\frac{5}{2}\\)y – \\(\\frac{9}{2}\\) = 0 ………. (ii)
\nS11<\/sub> = x\u00b2 + y\u00b2 + y = 0 ………. (iii)
\nRadical axis of S = 0, S1<\/sub> = 0 is S – S1<\/sub> = 0
\n\"Inter
\n5x – 5y – 5 = 0
\nx – y – 1 = 0 ………… (iv)
\nRadical axis of S = 0, S11<\/sub> = 0 is S – S11<\/sub> = 0
\n4x – y – 7 = 0 ……… (v)
\nx – y – 1 = 0………….. (vi)
\nSubtracting 3x – 6 = 0 \u21d2 3x = 6
\nx = \\(\\frac{6}{3}\\) = 2
\nSubstituting in (iv), 2 – y – 1 = 0
\ny = 1
\nRadical centre is P(2, 1)<\/p>\n

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

Question 1.
\nShow that the common chord of the circles x\u00b2 + y\u00b2 – 6x – 4y + 9 = 0 and x\u00b2 + y\u00b2 – 8x – 6y + 23 = 0 is the diameter of the second circle and also find its length.
\nSolution:
\nCommon chord be
\n(x\u00b2 + y\u00b2 – 6x – 4y + 9) – (x\u00b2 + y\u00b2 – 8x – 6y + 23) = 0
\n2x + 2y – 14 = 0
\nx + y – 7 = 0 ………… (i)
\nCentre of circle (-4, -3)
\n(-4, -3) lies on line x + y – 7
\nRadius is {4\u00b2 + 3\u00b2 – 23}\u00bd<\/sup> = \u221a2
\nDiameter = 2\u221a2<\/p>\n

Question 2.
\nFind the equation and length of the common chord of the following circles.
\ni) x\u00b2 + y\u00b2 + 2x + 2y + 1 =0,
\nx\u00b2 + y\u00b2 + 4x + 3y + 2 = 0.
\nSolution:
\nx\u00b2 + y\u00b2 + 2x + 2y + 1 = 0
\nx\u00b2 + y\u00b2 + 4x + 3y + 2 = 0
\nEquation of common chord is
\nS – S’ = 0 (x\u00b2 + y\u00b2 + 2x + 2y + 1) – (x\u00b2 + y\u00b2 + 4x + 3y + 2) = 0
\n-2x – y – 1 = 0
\n2x + y + 1 = 0
\nCentre of circle is (-1, -1)
\nRadius = \\(\\sqrt{1+1-1}\\) = 1
\nLength of \u22a5 from centre (-1, -1) to the chord is
\nd = \\(|\\frac{2(-1)+(-1)+1}{\\sqrt{2^{2}+1^{2}}}|=\\frac{2}{\\sqrt{5}}\\)
\n\"Inter<\/p>\n

ii) x\u00b2 + y\u00b2 – 5x – 6y + 4 = 0, x\u00b2 + y\u00b2 – 2x – 2 = 0
\nSolution:
\nCommon chord equation
\n(x\u00b2 + y\u00b2 – 5x- 6y + 4) – (x\u00b2+ y\u00b2 – 2x – 2) = 0
\n-3x – 6y + 6 = 0
\nx + 2y – 2 = 0
\n\"Inter<\/p>\n

Question 3.
\nProve that the radical axis of the circles x\u00b2 + y\u00b2 + 2gx + 2fy + c = 0 and x\u00b2 +y\u00b2 + 2g’x + 2f’y + c\u2018 = 0 is the diameter of the latter circle (or the former bisects the circumference of the latter) if 2g'(g – g’) + 2f (f – f’) – c – c’.
\nSolution:
\nRadical axis is
\n(x\u00b2 + y\u00b2 + 2gx + 2fy + c) – (x\u00b2 + y\u00b2 + 2g’x + 2fy + c’) = 0
\n2(g – g’) x + 2(f – f’)y + c – c’ = 0 …………. (i)
\nCentre of second circle is (-g1, -f1)
\nRadius = \\(\\sqrt{g’^{2}+f’^{2}+c’}\\)
\nNow (-g’, -f’) should lie on (i)
\n– 2g (g – g’) – 2f'(f – f’) + c – c’ = 0
\n(or) 2g (g – g’) + 2f'(f – f’) = c – c’<\/p>\n

Question 4.
\nShow that the circles x\u00b2 + y\u00b2 + 2ax + c = 0 and x\u00b2 + y\u00b2 + 2by + c = 0 touch each other if 1\/a\u00b2 + 1\/b\u00b2 = 1\/c.
\nSolution:
\nThe centres of the circles C1<\/sub> (-0, 0) and C2<\/sub> (0, -b) respectively
\nRadius of 1st circle be \\(\\sqrt{a^{2}-c}\\) = r1<\/sub>
\nRadius of 2nd circle be \\(\\sqrt{b^{2}-c}\\) = r2<\/sub>
\nC1<\/sub>C2<\/sub> = r1<\/sub> + r2<\/sub>
\n(C1<\/sub> C2<\/sub>)\u00b2 = (r1<\/sub> + r2<\/sub>)\u00b2
\n(a\u00b2 + b\u00b2) = a\u00b2 – c + b\u00b2 – c + 2\\(\\sqrt{a^{2}-c}.\\sqrt{a^{2}-c}\\)
\nc = \\(\\sqrt{a^{2}-c}.\\sqrt{a^{2}-c}\\)
\nc\u00b2 = (a\u00b2 – c) (b\u00b2 – c)
\nc\u00b2 = -c (a\u00b2 + b\u00b2) + a\u00b2b\u00b2 + c\u00b2
\n(or) c(a\u00b2 + b\u00b2) = a\u00b2b\u00b2 (or) \\(\\frac{1}{c}=\\frac{1}{a^{2}}+\\frac{1}{b^{2}}\\)<\/p>\n

\"Inter<\/p>\n

Question 5.
\nShow that the circles x\u00b2 + y\u00b2 – 2x = 0 and x\u00b2 + y\u00b2 + 6x – 6y + 2 = 0 touch each other. Find the coordinates of the point of contact. Is the point of contact external or internal?
\nSolution:
\nFor the circle S = x\u00b2 + y\u00b2 – 2x = 0
\nCentre C1<\/sub> = (1, 0) and Radius r1<\/sub> .= \\(\\sqrt{1+0}\\) = 1
\nFor the circle S’ = x\u00b2 + y\u00b2 + 6x – 6y + 2 = 0
\nCentre C2<\/sub> = (-3, 3) and
\n\"Inter
\nAs C1<\/sub>C2<\/sub> = r1<\/sub> + r2<\/sub> the two circles touch each other externally, the point of contact P divides line of centres internally in the ratio
\nr1<\/sub> : r2<\/sub> = 1 : 4
\nHence point of contact
\n\"Inter
\nThe contact of the circle is external.<\/p>\n

Question 6.
\nFind the equation of the circle which cuts the following circles orthogonally.
\ni) x\u00b2 + y\u00b2 + 4x – 7 = 0,
\n2x\u00b2 + 2y\u00b2 + 3x + 5y – 9 = 0,
\nx\u00b2 + y\u00b2 + y = 0.
\nSolution:
\nS = x\u00b2 + y\u00b2 + 4x – 7 = 0 ………… (i)
\nS1<\/sub> = 2x\u00b2 + 2y\u00b2 + 3x + 5y – 9 = 0
\n\"Inter
\nx – y – 1 = 0 ……. (iv)
\nRadical axis of S = 0, S11<\/sub> = 0 is S – S11<\/sub> = 0
\n4x – y – 7 = 0 ……….. (v)
\nSolving (iv) and (v)
\nWe get 3x – 6 = 0
\nx = -2
\nSubstute x value in (iv), 2 – y – 1 =0
\ny = 1
\nRadical centre is P(2, 1)
\nPT =’ Length of the tangent from P to S = 0
\n= \\(\\sqrt{4+1+8-7}\\)
\n= \u221a6
\nEquation of the circles cutting the given circles orthogonally
\n(x – 2)\u00b2 + (y – 1)\u00b2 = (\u221a6)\u00b2
\nx\u00b2 + 4 – 4x + y\u00b2 + 1 – 2y = 6
\nx\u00b2 + y\u00b2 – 4x – 2y – 1 = 0<\/p>\n

ii) x\u00b2 + y\u00b2 + 2x + 4y + 1 = 0,
\n2x\u00b2 + 2y\u00b2 + 6x + 8y – 3 = 0,
\nx\u00b2 + y\u00b2 – 2x + 6y – 3 = 0.
\nSolution:
\nEquations of the required circles are
\nS \u2261 x\u00b2 + y\u00b2 + 2x + 4y + 1 = 0
\nS1<\/sub> \u2261 x\u00b2 + y\u00b2 + 3x + 4y – \\(\\frac{3}{2}\\) = 0
\nS11<\/sub> \u2261 x\u00b2 + y\u00b2 – 2x + 6y – 3 = 0
\nRadical axis of S = 0, S1<\/sub> = 0 is S – S1<\/sub> =0
\n-x + \\(\\frac{5}{2}\\) = 0 \u21d2 x = \\(\\frac{5}{2}\\)
\nRadical axis of S = 0, S11<\/sub> = 0 is S – S11<\/sub> = 0
\n4x – 2y + 4 = 0
\n\u21d2 2x – y + 2 = 0
\nx = \\(\\frac{5}{2}\\) \u21d2 5 – y + 2 = 0
\n\u21d2 y = 7
\nRadical centre is P (\\(\\frac{5}{2}\\) , 7)
\nPT = Length of the tangent from P to S = 0
\n\"Inter
\nEquation of the circles cutting the given circles orthogonally
\n\"Inter<\/p>\n

iii) x\u00b2 + y\u00b2 + 2x + 17y + 4 = 0,
\nx\u00b2 + y\u00b2 + 7x + 6y + 11 = 0,
\nx\u00b2 + y\u00b2 – x + 22y + 3 = 0
\nSolution:
\nEquations of the required circles are
\nS \u2261 x\u00b2 + y\u00b2 + 2x + 17y + 4 = 0 ……… (i)
\nS1<\/sub> \u2261 x\u00b2 + y\u00b2 + 7x + 6y + 11 = 0 ……… (ii)
\nS11<\/sub> \u2261 x\u00b2 + y\u00b2 – x + 22y + 3 = 0 ……… (iii)
\nRadical axis of S = 0, S1<\/sub> = 0 is S – S1<\/sub> = 0
\n-5x + 11 y – 7 = 0
\n5x – 11y + 7 = 0 ……… (iv)
\nRadical axis of S = 0, S11<\/sub> = 0 is S – S11<\/sub> =0
\n3x – 5y + 1 = 0 ………… (v)
\nSolving (iv) and (v)
\n\"Inter
\nRadical centre is P(3, 2)
\nPT = Length of the tangeqt from P to S = 0
\n= \\(\\sqrt{9+4+6+34+4}\\) = \u221a57
\nEquation of the circle cutting the given circles cutting orthogonally
\n(x – 3)\u00b2 + (y – 2)\u00b2 = 57
\nx\u00b2 – 6x + 9 + y\u00b2 – 4y + 4 – 57 = 0
\nx\u00b2 + y\u00b2 – 6x – 4y – 44 = 0<\/p>\n

\"Inter<\/p>\n

iv) x\u00b2 + y\u00b2 + 4x + 2y + 1 = 0,
\n2(x\u00b2 + y\u00b2) + 8x + 6y – 3 = 0,
\nx\u00b2 + y\u00b2 + 6x – 2y – 3 = 0.
\nSolution:
\nEquations of the required circles are
\nS \u2261 x\u00b2 + y\u00b2 + 4x + 2y + 1 = 0 ……….. (i)
\nS1<\/sub> \u2261 x\u00b2 + y\u00b2 + 4x + 3y – \\(\\frac{3}{2}\\) = 0 ………… (ii)
\nS11<\/sub> \u2261 x\u00b2 + y\u00b2 + 6x – 2y – 3 = 0 ………. (iii)
\n(i) – (ii) gives radical axis of S = 0, S1<\/sub> = 0 is
\nS – S1<\/sub> = 0 \u21d2 -y + \\(\\frac{5}{2}\\) = 0 \u21d2 y = \\(\\frac{5}{2}\\)
\nRadical axis of S = 0, S11<\/sub> = 0 is S – S11<\/sub> = 0
\n– 2x + 4y + 4 = 0
\nx – 2y – 2 = 0
\ny = \\(\\frac{5}{2}\\) \u21d2 x – 5 – 2 = 0
\nx = 5 + 2 = 7
\nRadical centre is P (7, \\(\\frac{5}{2}\\))
\nPT = Length of the tangent P to S = 0
\n\"Inter
\nEquation of the required circle is
\n\"Inter<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B System of Circles Solutions Exercise 2(b) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2B System of Circles Solutions Exercise 2(b) I. Question 1. Find the equation of the radical axis of the following circles, i) x\u00b2 …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35514"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35514"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35514\/revisions"}],"predecessor-version":[{"id":35515,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35514\/revisions\/35515"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35514"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35514"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35514"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}