AP EAMCET · Maths · Circle
The inverse point of \((1,3)\) with respect to the circle \(x^2+y^2-4 x-6 y+9=0\) is
- A \((2,3)\)
- B \((2,-3)\)
- C \((-2,3)\)
- D \((-2,-3)\)
Answer & Solution
Correct Answer
(C) \((-2,3)\)
Step-by-step Solution
Detailed explanation
Center \(C=(-(-4)/2, -(-6)/2) = (2,3)\). \(r^2 = (-2)^2 + (-3)^2 - 9 = 4+9-9 = 4\). Let \(P=(1,3)\) and inverse point \(P'=(x',y')\). \(CP^2 = (1-2)^2+(3-3)^2 = (-1)^2+0^2 = 1\). \(x' = 2 + \frac{4(1-2)}{1} = 2 - 4 = -2\). \(y' = 3 + \frac{4(3-3)}{1} = 3 + 0 = 3\). Inverse…
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