Question

Let O be the centre of the circle x²+y²=r², where r > √5/2. 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 _____

Answer 1

Method -1 :-

From the figure,

➠ OA = √5 /2   & OC  = 4/√5

➠CQ = OC = 4/√5  & CA = 3/2√5.

By pythagoras theorem,

➠ OQ = √OA² + AQ²

        = OA²+(CQ²-CA²)

➠ √ 5/4 + 16/5 - 9/20  = √4 = 2.

Hence,$\boxed{ r = 2.}$

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Method -2:-

From the figure ,

➠  PQ is  hx +ky = r²

Given that,

➠ PQ  : 2x +4y = 5

Therefore,

➠  h/2 = k/4 = r² /5

➠ h = 2r² / 5  &  k = 4r²/5

∴ C = (r² /5 , 2r²/5 )

Hence, C lies on  x+2y = 4

So,

➠ r² /5 + 2 (2r² /5 ) = 4

➠ r² = 4

➠ r = 2.

_______________________

➥ $\pink{\boxed{Value\ of \  r = 2.}}$

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Answer 2

ANSWER.

The value of r is = 2.

EXPLANATION.

$\sf : \implies \: origin \: = (0,0) \: \: parallel \: to \: the \: line \: = 2x + 4y = 5 \\ \\ \sf : \implies \: by \: using \: distance \: formula \\ \\ \sf : \implies \: om \: = | \frac{2x + 4y - 5}{ \sqrt{ {2}^{2} + {4}^{2} } } | \\ \\ \sf : \implies \: om \: = | \frac{2(0) + 4(0) - 5}{ \sqrt{20} } | = \frac{5}{2 \sqrt{5} }$

$\sf : \implies \: oc \: = | \dfrac{x + 2y - 4}{ \sqrt{ {1}^{2} + {2}^{2} } } | \\ \\ \sf : \implies \: oc \: = | \frac{0 + 2(0) - 4}{ \sqrt{5} } | = \frac{4}{ \sqrt{5} } \\ \\ \sf : \implies \: it \: is \: also \: written \: as \: = \frac{8}{2 \sqrt{5} }$

$\sf : \implies \: cm \: = oc \: - \: om \\ \\ \sf : \implies \: cm \: = \frac{8}{2 \sqrt{5} } - \frac{5}{2 \sqrt{5} } = \frac{3}{2 \sqrt{5} } \\ \\ \sf : \implies \: {cp}^{2} = {co}^{2} \\ \\ \sf : \implies \: {cm}^{2} + {pm}^{2} = {co}^{2}$

$\sf : \implies \: \dfrac{9}{20} + {r}^{2} - {om}^{2} = \dfrac{16}{5} \\ \\ \sf : \implies \: \: \frac{9}{20} + {r}^{2} - \frac{5}{4} = \frac{16}{5} \\ \\ \sf : \implies \: {r}^{2} = \frac{16}{5} + \frac{5}{4} - \frac{9}{20} \\ \\ \sf : \implies \: {r}^{2} = \frac{64 + 25 - 9}{20} = \frac{80}{20} \\ \\ \sf : \implies \: {r}^{2} = 4 \: \: \: and \: \: \: r \: = 2$