TS EAMCET · Maths · Circle
The equation of the circle whose radius is 3 and which touches the circle \(x^2+y^2-4 x-6 y-12=0\) internally at \((-1,-1)\) is
- A \(5 x^2+5 y^2-8 x-14 y-32=0\)
- B \(x^2+y^2-12 x-14 y-28=0\)
- C \(3 x^2+3 y^2-8 x-14 y-31=0\)
- D \(x^2+y^2-5 x-7 y-14=0\)
Answer & Solution
Correct Answer
(A) \(5 x^2+5 y^2-8 x-14 y-32=0\)
Step-by-step Solution
Detailed explanation
Center of given circle: \(O_1 = (2, 3)\). Radius of given circle: \(r_1 = \sqrt{2^2+3^2-(-12)} = 5\). Point of tangency: \(P = (-1, -1)\). Radius of required circle: \(r_2 = 3\). Since the circles touch internally, the center \(O_2\) of the required circle divides the line…
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