AP EAMCET · Maths · Circle
If the equation of the circle whose radius is 3 units and which touches internally the circle \(x^2+y^2-4 x-6 y-12\) \(=0\) at the point \((-1,-1)\) is \(x^2+y^2+p x+q y+r=0\), then \(p+q-r=\)
- A 2
- B \(\frac{5}{2}\)
- C \(\frac{26}{5}\)
- D 3
Answer & Solution
Correct Answer
(A) 2
Step-by-step Solution
Detailed explanation
\(x^2+y^2+p x+q y+r=0\) touches the given circle at \((-1,-1)\). Hence \((-1)^2+(-1)^2-p-q+r=0\) \(\Rightarrow p+q-r=2\).
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