MHT CET · Maths · Circle
If the radius of a circle \(x^{2}+y^{2}-4 x+6 y-k=0\) is 5, then \(\mathrm{k}=\)
- A \(-12\)
- B \(-25\)
- C \(25\)
- D \(12\)
Answer & Solution
Correct Answer
(D) \(12\)
Step-by-step Solution
Detailed explanation
Given equation of circle is
\(
\begin{array}{l}
x^{2}+y^{2}-4 x+6 y-k=0 \\
r \quad=\sqrt{4+9+k} \Rightarrow 5=\sqrt{13+k} \Rightarrow 13+k=25 \\
k=12
\end{array}
\)
\(
\begin{array}{l}
x^{2}+y^{2}-4 x+6 y-k=0 \\
r \quad=\sqrt{4+9+k} \Rightarrow 5=\sqrt{13+k} \Rightarrow 13+k=25 \\
k=12
\end{array}
\)
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