Circle Ques 1
- Let $C_1$ and $C_2$ be the centres of the circles $x^2+y^2-2 x-2 y-2=0$ and $x^2+y^2-6 x-6 y+14=0$ respectively. If $P$ and $Q$ are the points of intersection of these circles, then the area (in sq units) of the quadrilateral $P C_1 Q C_2$ is
(2019 Main, 12 Jan I)
$8$
$4$
$6$
$9$
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Answer:
Correct Answer: 1.(b)
Solution: (b) Given circles, the number of common tangents depends on their relative positions. If they are separate, there are 4 common tangents; if they are externally tangent, 3; if intersecting, 2; if internally tangent, 1; and if one is inside the other without touching, 0.
$ x^2+y^2-2 x-2 y-2=0 $
and
$ x^2+y^2-6 x-6 y+14=0 $
are intersecting each other orthogonally, since
$ 2(1)(3)+2(1)(3)=12-2 $
$[\because$ two circles intersect orthogonally if $\left.2 g_1 g_2+2 f_1 f_2=c_1+c_2\right]$
So, area of quadrilateral is calculated by dividing it into triangles or using specific formulas based on its type.
$ \begin{alignedat} P C_1 Q C_2 & =2 \times \operatorname{ar}\left(\triangle P C_1 C_2\right) . \\ & =2 \times\left(\frac{1}{2} \times 2 \times 2\right)=4 \text { sq units } \end{aligned} $
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