TS EAMCET · Maths · Circle
The normal to the circle given by \(x^2+y^2-6 x+8 y-144=0\) at \((8,8)\) meets the circle again at the point
- A \((2,-16)\)
- B \((2,16)\)
- C \((-2,16)\)
- D \((-2,-16)\)
Answer & Solution
Correct Answer
(D) \((-2,-16)\)
Step-by-step Solution
Detailed explanation
Given equation of circle, \[ x^2+y^2-6 x+8 y-144=0 \] Comparing with \(x^2+y^2+2 g x+2 f y+c=0\) \[ \begin{aligned} 2 g & =-6 \Rightarrow g=-3 \\ 2 f & =8 \Rightarrow f=4 \end{aligned} \] Centre \((-g,-f)=(3,-4)\) We know that, normal to a circle passes through its centre. Let…
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