AP EAMCET · Maths · Circle
\(3 x+4 y-43=0\) is a tangent to the circle \(S \equiv x^2+y^2-6 x+8 y+k=0\) at a point P. If C is the centre of the circle and Q is a point which divides CP in the ratio \(-1: 2\), then the power of the point Q with respect to the circle \(\mathrm{S}=0\) is
- A \(50\)
- B \(21\)
- C \(0\)
- D \(5\)
Answer & Solution
Correct Answer
(C) \(0\)
Step-by-step Solution
Detailed explanation
Center of circle \(C = (3, -4)\). Radius \(r = \frac{|3(3)+4(-4)-43|}{\sqrt{3^2+4^2}} = \frac{|9-16-43|}{\sqrt{25}} = \frac{|-50|}{5} = 10\). Point P (tangency) is the foot of the perpendicular from C to \(3x+4y-43=0\). Line CP:…
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