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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 \(p+q-r=\)

  1. A 2
  2. B \(\frac{5}{2}\)
  3. C \(\frac{26}{5}\)
  4. D 3
Verified Solution

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 \((-1)^2+(-1)^2-p-q+r=0\) \(\Rightarrow p+q-r=2\).