AP EAMCET · Maths · Circle
The equation of the circle whose radius is 3 and which touches internally the circle \(x^2+y^2-4 x-6 y-12=0\) at the point \((-1,-1)\) is
- A \(5 x^2+5 y^2+9 x-6 y-7=0\)
- B \(5 x^2+5 y^2-8 x-14 y-32=0\)
- C \(5 x^2+5 y^2-6 x+8 y-8=0\)
- D \(5 x^2+5 y^2+6 x-8 y-12=0\)
Answer & Solution
Correct Answer
(B) \(5 x^2+5 y^2-8 x-14 y-32=0\)
Step-by-step Solution
Detailed explanation
Equation of given circle is \(x^2+y^2-4 x-6 y-12=0\) having centre \(C_1(2,3)\) and radius \(r_1=\sqrt{4+9+12}=5\). Let the required circle having centre \(C_2(h, k)\) and radius is given as 3 touches the given circle at \(A(-1,-1)\). The point \(A(-1,-1)\) divides the line…
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