TS EAMCET · Maths · Circle
The equation of the circle passing through the points of intersection of the two orthogonal circles \(\begin{aligned} & S_1=x^2+y^2+k x-4 y-1=0, \ & S_2=3 x^2+3 y^2-14 x+23 y-15=0 \text { and passing } \end{aligned}\) through the point \((-1,-1)\) is
- A \(x^2+y^2-8 x-2 y-12=0\)
- B \(3 x^2+3 y^2+18 x-12 y=0\)
- C \(5 x^2+5 y^2-22 x+15 y-17=0\)
- D \(x^2+y^2-5 x+14 y+7=0\)
Answer & Solution
Correct Answer
(C) \(5 x^2+5 y^2-22 x+15 y-17=0\)
Step-by-step Solution
Detailed explanation
We have, \(\begin{aligned} S_1: x^2+y^2+k x-4 y-1 & =0 \\ S_2: 3 x^2+3 y^2-14 x+23 y-15 & =0 \\ =x^2+y^2-\frac{14}{3} x+\frac{23}{3} y-5 & =0 \end{aligned}\) Since, \(S_1\) and \(S_2\) are orthogonal…
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