MHT CET · Maths · Circle
The equation of a circle, which passes through the centre of the circle \(x^2+y^2+8 x+10 y-7=0\) and is concentric with the circle. \(2 x^2+2 y^2-8 x-12 y-9=0\), is
- A \(x^2+y^2-4 x+6 y-87=0\)
- B \(x^2+y^2+4 x+6 y-87=0\)
- C \(x^2+y^2+4 x+6 y+87=0\)
- D \(x^2+y^2-4 x-6 y-87=0\)
Answer & Solution
Correct Answer
(D) \(x^2+y^2-4 x-6 y-87=0\)
Step-by-step Solution
Detailed explanation
The required circle is concentric with the circle
\(
2 x^2+2 y^2-8 x-12 y-9=0
\)
i.e, centre is at \((2,3)\) and passes through the centre of the circle
\(
x^2+y^2+8 x+10 y-7=0
\)
i.e., through \((-4,-5)\)
Hence, the required equation is
\(
\begin{aligned}
& (x-2)^2+(y-3)^2=(2+4)^2+(3+5)^2 \\
& \Rightarrow x^2+y^2-4 x-6 y-87=0
\end{aligned}
\)
\(
2 x^2+2 y^2-8 x-12 y-9=0
\)
i.e, centre is at \((2,3)\) and passes through the centre of the circle
\(
x^2+y^2+8 x+10 y-7=0
\)
i.e., through \((-4,-5)\)
Hence, the required equation is
\(
\begin{aligned}
& (x-2)^2+(y-3)^2=(2+4)^2+(3+5)^2 \\
& \Rightarrow x^2+y^2-4 x-6 y-87=0
\end{aligned}
\)
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