A circle C touches the line x = 2y at the point (2, 1) and intersects the circle
C1 : x2 + y2 + 2y $-$ 5 = 0 at two points P and Q such that PQ is a diameter of C1. Then the diameter of C is :
Solution
(x $-$ 2)<sup>2</sup> + (y $-$ 1)<sup>2</sup> + $\lambda$(x $-$ 2y) = 0<br><br>C : x<sup>2</sup> + y<sup>2</sup> + x($\lambda$ $-$ 4) + y($-$2 $-$2$\lambda$) + 5 = 0<br><br>C<sub>1</sub> : x<sup>2</sup> + y<sup>2</sup> + 2y $-$ 5 = 0<br><br>S<sub>1</sub> $-$ S<sub>2</sub> = 0 (Equation of PQ)<br><br>($\lambda$ $-$ 4)x $-$ (2$\lambda$ + 4)y + 10 = 0 Passes through (0, $-$1)<br><br>$\Rightarrow$ $\lambda$ = $-$7<br><br>C : x<sup>2</sup> + y<sup>2</sup> $-$ 11x + 12y + 5 = 0<br><br>= ${{\sqrt {245} } \over 4}$<br><br>Diameter = $7\sqrt 5$
About this question
Subject: Mathematics · Chapter: Circles · Topic: Equation of a Circle
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