MHT CET · Maths · Circle
The equation of the circle, the end points of whose diameter are the centres of the circles \(x^2+y^2+6 x-14 y+5=0\) and \(x^2+y^2-4 x+10 y-4=0\) is
\(x^2+y^2-4 x+10 y-4=0\) is
- A \(x^2+y^2-x-2 y+41=0\)
- B \(x^2+y^2+x-2 y-41=0\)
- C \(x^2+y^2+x-2 y-41=0\)
- D \(x^2+y^2-x+2 y-41=0\)
Answer & Solution
Correct Answer
(C) \(x^2+y^2+x-2 y-41=0\)
Step-by-step Solution
Detailed explanation
Center of circle \(x^2+y^2+6 x-14 y+5=0\) is \((-3,7)\) and centre of circle \(x^2+y^2-4 x+10 y-4=0\) is \((2,-5)\)
\(\therefore \quad\) Equation of the required circle is
\(\begin{aligned}
& (x+3)(x-2)+(y-7)(y+5)=0 \\
& \Rightarrow x^2+y^2+x-2 y-41=0
\end{aligned}\)
\(\therefore \quad\) Equation of the required circle is
\(\begin{aligned}
& (x+3)(x-2)+(y-7)(y+5)=0 \\
& \Rightarrow x^2+y^2+x-2 y-41=0
\end{aligned}\)
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