TS EAMCET · Maths · Circle
If \(P\) is the point of contact of the circles \(x^2+y^2+4 x+4 y-10=0\) and \(x^2+y^2-6 x-6 y+10=0\) and \(Q\) is their external centre of similitude, then the equation of the circle with \(P\) and \(Q\) as the extremities of its diameter is
- A \(x^2+y^2+14 x+14 y-26=0\)
- B \(x^2+y^2+5 x+5 y-8=0\)
- C \(x^2+y^2-5 x-5 y+8=0\)
- D \(x^2+y^2-14 x-14 y+26=0\)
Answer & Solution
Correct Answer
(D) \(x^2+y^2-14 x-14 y+26=0\)
Step-by-step Solution
Detailed explanation
We have, \[ \begin{aligned} & x^2+y^2+4 x+4 y-10=0 \text { and } \\ & x^2+y^2-6 x-6 y+10=0 \end{aligned} \] Centre \[ \begin{aligned} & c_1=(-2,-2) \text { and } r_1=3 \sqrt{2} \\ & c_2=(3,3) \text { and } r_2=2 \sqrt{2} \end{aligned} \] \(P\) is the point of contact of circle…
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