TS EAMCET · Maths · Circle
The equation of a circle which passes through the points of intersection of the circles \(2 x^2+2 y^2-2 x+6 y-3=0\), \(x^2+y^2+4 x+2 y+1=0\) and whose centre lies on the common chord of these circles is
- A \(2 x^2+2 y^2-3 x+4 y-2=0\)
- B \(x^2+y^2+2 x+5 y-2=0\)
- C \(3 x^2+3 y^2-2 x+4 y-3=0\)
- D \(4 x^2+4 y^2+6 x+10 y-1=0\)
Answer & Solution
Correct Answer
(D) \(4 x^2+4 y^2+6 x+10 y-1=0\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & S_1 \equiv 2 x^2+2 y^2-2 x+6 y-3=0 \\ & \Rightarrow x^2+y^2-x+3 y-\frac{3}{2}=0 \\ & S_2 \equiv x^2+y^2+4 x+2 y+1=0\end{aligned}\) Eq of common chord is \(S_1-S_2=0\) i.e. \(-5 x+y-\frac{5}{2}=0\)..(i) Eq. of circle passing through intersections of \(S_1\)…
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