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Q.

Consider the circle $x^{2} + y^{2} - 4 x - 2 y + c = 0$ whose centre is A(2,1). If the point P(10,7) is such that the line segment PA meets the circle in Q with PQ = 5 then c =

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a

$- 20$

b

$- 15$

c

20

d

30

answer is D.

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Detailed Solution

$G i v e n c i r c l e i s x^{2} + y^{2} - 4 x - 2 y + c = 0 \Rightarrow {(x - 2)}^{2} + {(y - 1)}^{2} = 5 - c N o w P A = \sqrt{64 + 36 } = 10 P Q = 5 (G i v e n) ∴ A Q = 10 - 5 = 5 ∴ R a d i u s = 5 u n i t S o, 5 - c = 25 \Rightarrow c = - 20$

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Consider the circle x2+y2-4x-2y+c=0 whose centre is A(2,1). If the point P(10,7) is such that the line segment PA meets the circle in Q with PQ = 5 then c =