Test Index

Circles Part 2

Section: Mathematics

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Question : 53 of 87

Marks: +1, -0

If the tangents at the points

P

Q

x^2+y^2-2 x+y=5

meet at the point

R \left( \frac{9}{4}, 2 \right)

, then the area of the triangle

PQR

[6-Apr-2023 shift 2]

$\; \frac{5}{4}$
$\; \frac{13}{4}$
$\; \frac{5}{8}$
$\; \frac{13}{8}$

Solution:

x^2+y^2-2 x+y=5

with resperct to

R P Q

is

x\; \text{ C.O.C }\;

eq^{n}

of C.O.C is

x_{T=0}

\;\; \frac{9}{4} x + 2 y - \left( x + \frac{9}{4} \right) + \frac{1}{2}(y+2) - 5 = 0

\;\; \frac{5}{4} x + \frac{5}{2} y - \frac{25}{4} = 0

\; 5 x+10 y-25=0

\; x+2 y=5

\;\; \text{ Area }\;\; = \frac{1}{2} \left( P' \right) \left( PQ \right) \;\;\; (PQ)=2 \sqrt{ r^2 - p^2 } = \sqrt{5}

= \frac{1}{2} \left[ \frac{\sqrt{5}}{4} \right] \left( \sqrt{5} \right) \;\;\; P' = \frac{ \frac{9}{4} + 4 - 5 }{ \sqrt{5} }

\;\; = \frac{5}{8} \;\;\; = \left( \frac{5}{4 \sqrt{5}} \right) = \frac{\sqrt{5}}{4}

Method II

\;\; \text{ area }\; = \frac{ (RL)^3 }{ R^2 + L^2 }

\; R = \frac{5}{2}

\; L = \sqrt{ \frac{81}{16} + 4 - \frac{9}{2} + 2 - 5 }

\; = \frac{5}{4}

\;\; \text{ area }\; == \frac{5}{8}

Ans. Option 3

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