Let the tangent to the circle x2+y2=25 at the point R (3,4) meet x -axis and y -axis at point P and Q, respectively. If r is the radius of the circle passing through the origin O and having centre at the incentre of the triangle OPQ, then r 2 is equal to

Question

Let the tangent to the circle x2+y2=25 at the point R (3,4) meet x -axis and y -axis at point P and Q, respectively. If r is the radius of the circle passing through the origin O and having centre at the incentre of the triangle OPQ, then r 2 is equal to (A) (529/64) (B) (125/72) (C) (625/72) (D) (585/66)

Tardigrade Admin · Accepted Answer

Tangent to circle 3x + 4y = 25 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/a9cbad5859c0ebc0ebd6b01af684db8d-.png" /> OP + OQ + OR =25 Incentre =(((25/4) × (25/3)/25), ((25/4) × (25/3)/25)) =((25/12), (25/12)) ∴ r 2=2((25/12))2=2 × (625/144)=(625/72)

Q. Let the tangent to the circle $x^{2} + y^{2} = 25$ at the point $R (3, 4)$ meet $x$ -axis and $y$ -axis at point $P$ and $Q$ , respectively. If $r$ is the radius of the circle passing through the origin $O$ and having centre at the incentre of the triangle $OPQ$ , then $r^{2}$ is equal to

$\frac{529}{64}$

$\frac{125}{72}$

$\frac{625}{72}$

$\frac{585}{66}$

Tangent to circle $3 x + 4 y = 25$

$OP + OQ + OR = 25$
Incentre $= (\frac{\frac{25}{4} \times \frac{25}{3}}{25}, \frac{\frac{25}{4} \times \frac{25}{3}}{25})$
$= (\frac{25}{12}, \frac{25}{12})$
$∴ r^{2} = 2 (\frac{25}{12})^{2} = 2 \times \frac{625}{144} = \frac{625}{72}$

Q. Let the tangent to the circle x2+y2=25 at the point R(3,4) meet x -axis and y -axis at point P and Q, respectively. If r is the radius of the circle passing through the origin O and having centre at the incentre of the triangle OPQ, then r2 is equal to

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