Let B be centre of circle x^{2}+y^{2}-2 x+4 y+1=0 Let tangents at two points mathrm{P} and mathrm{Q}

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10-1.Circle and System of Circles

hard

Let $B$ be the centre of the circle $x^{2}+y^{2}-2 x+4 y+1=0$ Let the tangents at two points $\mathrm{P}$ and $\mathrm{Q}$ on the circle intersect at the point $\mathrm{A}(3,1)$. Then $8.$ $\left(\frac{\text { area } \triangle \mathrm{APQ}}{\text { area } \triangle \mathrm{BPQ}}\right)$ is equal to .... .

$18$

$36$

$72$

$12$

(JEE MAIN-2021)

Solution

$\tan \theta=\frac{3}{2}$

$\frac{\text { Area } \Delta \mathrm{APQ}}{\text { Area } \Delta \mathrm{BPQ}}=\frac{\mathrm{AR}}{\mathrm{RB}}=\frac{3 \sin \theta}{2 \cos \theta}=\frac{9}{4}$

$8\left(\frac{\text { Area } \Delta \mathrm{APQ}}{\text { Area } \Delta \mathrm{BPQ}}\right)=18$

Standard 11

Mathematics

A pair of tangents are drawn to a unit circle with centre at the origin and these tangents intersect at A enclosing an angle of $60^o$. The area enclosed by these tangents and the arc of the circle is

normal

Two tangents are drawn from the point $\mathrm{P}(-1,1)$ to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-6 \mathrm{y}+6=0$. If these tangents touch the circle at points $A$ and $B$, and if $D$ is a point on the circle such that length of the segments $A B$ and $A D$ are equal, then the area of the triangle $A B D$ is eqaul to:

medium

(JEE MAIN-2021)

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easy

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hard

Let $O$ be the origin and $OP$ and $OQ$ be the tangents to the circle $x^2+y^2-6 x+4 y+8=0$ at the point $P$ and $Q$ on it. If the circumcircle of the triangle OPQ passes through the point $\left(\alpha, \frac{1}{2}\right)$, then a value of $\alpha$ is

hard

(JEE MAIN-2023)

Solution

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