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Let $C_{1}\: \: and\: \: C_{2}$ be the centres of the circles $x^{2}+y^{2}-2x-2y-2=0$ and $x^{2}+y^{2}-6x-6y+14=0$ respectively. If P and Q are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral $PC_{1}QC_{2}$ is :
Option 1)
$8$
Option 2)
$4$
Option 3)
$9$
Option 4)
$6$

Answers (1)

best_answer

@5089

General form of a circle -

$x^{2}+y^{2}+2gx+2fy+c= 0$

- wherein

centre = $\left ( -g,-f \right )$

radius = $\sqrt{g^{2}+f^{2}-c}$

Common tangents of two circle -

When they intersect, there are two common tangents, both of them being direct.

$|r_1-r_2|<\left|C_1C_2\right|<r_1+r_2$

- wherein

Equation of circle with centre $C_{1}(1,1)$

$(x-1)^{2}+(y-1)^{2}=4$

Equation of circle with centre $C_{2}(3,3)$

$(x-3)^{2}+(y-3)^{2}=4$

Two circles are orthogonal ,

Hence ,

Area = $2(\frac{1}{2}\cdot 2\cdot 2)=4 \: sq. units$

Option 1)

$8$

Option 2)
$4$
Option 3)

$9$

Option 4)

$6$

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