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  • Find the centres of the circles which pass through the ends of the common chords of two circles <img alt="2 x^2+2 y^2+8 x+4 y-7=0" src="https://entrancecorner.oncodecogs.com/gif.latex?2%20x%5E2&amp

Find the centres of the circles which pass through the ends of the common chords of two circles 2 x^2+2 y^2+8 x+4 y-7=0 and x^2+y^2-8 x-4 y-5=0and touch the line x=7.

Option: 1

\mathrm{(12,6) \text { and }(-60,-30)}


Option: 2

\mathrm{(12,3) \text { and }(-60,30)}


Option: 3

\mathrm{\left(3, \frac{3}{2}\right) \text { and }(-60,-30)}


Option: 4

\mathrm{(6,3) \text { and }(60,-30)}


Answers (1)

best_answer

The equation to the common chord of the two given circles is

2 x^2+2 y^2+8 x+4 y-7-2\left(x^2+y^2-8 x-4 y-5\right)=0 \text { i.e., } 24 x+12 y+3=0 \text { i.e., } 8 x+4 y+1=0

Any circle through the points of intersection of this line and the circle 

x^2+y^2-8 x-4 y-5=0 \text {, is of the form } x^2+y^2-8 x-4 y-5+\lambda(8 x+4 y+1)=0

This meets x = 7, at points whose ordinates are given by

49+y^2-56-4 y-5+\lambda(56+4 y+1)=0 \text { i.e., } y^2+4(\lambda-1) y+57 \lambda-12=0

This quadratic has equal roots since x = 7 is to be a tangent.

16(\lambda-1)^2-4(57 \lambda-12)=0 \text {, i.e., } 4 \lambda^2-65 \lambda+16=0 \Rightarrow(4 \lambda-1)(\lambda-16)=0

\lambda=\frac{1}{4} ; \lambda=16

For  \lambda=\frac{1}{4} one circle is x^2+y^2-8 x-4 y-5+\frac{1}{4}(8 x+4 y+1)=0

\\ i.e, 4\left(x^2+y^2\right)-24 x-12 y-19=0

For \lambda=16,2^{\text {nd }} circle is x^2+y^2-8 x-4 y-5+16(8 x+4 y+1)=0

i.e, \quad x^2+y^2+120 x+60 y+11=0

 

Posted by

Ajit Kumar Dubey

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