KCET · Maths · Circle
The number of real circles cutting orthogonally the circle \(x^{2}+y^{2}+2 x-2 y+7=0\) is
- A 0
- B 1
- C 2
- D Infinitely many
Answer & Solution
Correct Answer
(A) 0
Step-by-step Solution
Detailed explanation
Given, equation of circle is
\(x^{2}+y^{2}+2 x-2 y+7=0\)
Here, radius of the circle
\(\begin{aligned}
&=\sqrt{(1)+(-1)^{2}-7} \\
&=\sqrt{1+1-7}=\sqrt{-5} \\
&=\text { imaginary }
\end{aligned}\)
\(\therefore\) Given circle is an imaginary circle.
Hence, number of real circles cutting orthogonally the given imaginary circle is zero.
\(x^{2}+y^{2}+2 x-2 y+7=0\)
Here, radius of the circle
\(\begin{aligned}
&=\sqrt{(1)+(-1)^{2}-7} \\
&=\sqrt{1+1-7}=\sqrt{-5} \\
&=\text { imaginary }
\end{aligned}\)
\(\therefore\) Given circle is an imaginary circle.
Hence, number of real circles cutting orthogonally the given imaginary circle is zero.
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