Let be the centre of the circle , where . Suppose is a chord of this circle and the equation of the line passing through P and Q is . If the centre of the circumcircle of the triangle OPQ lies on the line , then the value of is _____ .

Question

Let $O$ be the centre of the circle $x^{2} + y^{2} = r^{2}$ , where $r > \frac{\sqrt{5}}{2}$ . Suppose $P Q$ is a chord of this circle and the equation of the line passing through P and Q is $2 x + 4 y = 5$ . If the centre of the circumcircle of the triangle OPQ lies on the line $x + 2 y = 4$ , then the value of $r$ is _____ . [2020]

Smart Achievers · Accepted Answer

(2)

$∵$ $Centre of circle is O (0, 0)$

$O A = perpendicular distance from point O to line$

$2 x + 4 y = 5$ $=$ $| \frac{0 + 0 - 5}{\sqrt{4 + 16}} |$ $= \frac{\sqrt{5}}{2}$

$O C = perpendicular distance from point O to line$

$x + 2 y = 4$ $=$ $| \frac{0 + 0 - 4}{\sqrt{1 + 4}} |$ $= \frac{4}{\sqrt{5}}$

$∴$ $C A = O C - O A = \frac{3}{2 \sqrt{5}}$ $∵$ $C Q = O C = \frac{4}{\sqrt{5}} (radius)$

$Now A Q^{2} = C Q^{2} - C A^{2} (∵ A C ⊥ P Q) = \frac{16}{5} - \frac{9}{20} = \frac{11}{4}$

$∴$ $O Q = r = \sqrt{O A^{2} + A Q^{2}}$

$\Rightarrow r = \sqrt{\frac{5}{4} + \frac{11}{4}} = \sqrt{4} = 2$

Solution

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Solution

Q. Let O be the centre of the circle x2+y2=r2, where r>52. Suppose PQ is a chord of this circle and the equation of the line passing through P and Q is 2x+4y=5. If the centre of the circumcircle of the triangle OPQ lies on the line x+2y=4, then the value of r is _____ . [2020]

Ans. (2) ∵ Centre of circle is O(0,0) OA=perpendicular distance from point O to line 2x+4y=5=|0+0-54+16|=52 OC=perpendicular distance from point O to line x+2y=4=|0+0-41+4|=45 ∴ CA=OC-OA=325 ∵ CQ=OC=45 (radius) Now AQ2=CQ2-CA2 (∵AC⊥PQ)=165-920=114 ∴ OQ=r=OA2+AQ2 ⇒r=54+114=4=2

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