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If the incentre of an equilateral triangle is (1, 1) and the equation of its one side is $3x+4y+3=0,$

then the equation of the circumcircle of this triangle is :

Option 1)
$x^{2}+y^{2}-2x-2y-2=0$

Option 2)
$x^{2}+y^{2}-2x-2y-14=0$

Option 3)
$x^{2}+y^{2}-2x-2y+2=0$

Option 4)
$x^{2}+y^{2}-2x-2y-7=0$

Answers (1)

As learnt in concept

Perpendicular distance of a point from a line -

$\rho =\frac{\left | ax_{1}+by_{1}+c\right |}{\sqrt{a^{2}+b^{2}}}$

- wherein

$\rho$ is the distance from the line $ax+by+c=0$ .

Equation of a circle -

$\left ( x-h \right )^{2}+\left ( y-k \right )^{2}= r^{2}$

- wherein

Circle with centre $\left ( h,k \right )$ and radius $r$ .

Distance of O from BC

is $\frac{\left | 3\left ( 1 \right )+4\left ( 1 \right )+3 \right |}{5}$

= 2 units

$Also \: \frac{AO}{DO}=\frac{2}{1}=\frac{AO}{2}$

$=> AO= 4 \: units$

Hence circumcircle has r = 4 and centre (1,1)

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

$x^{2}+y^{2}-2x-2y-14=0$

Option 1)

$x^{2}+y^{2}-2x-2y-2=0$

Incorrect option

Option 2)

$x^{2}+y^{2}-2x-2y-14=0$

Correct option

Option 3)

$x^{2}+y^{2}-2x-2y+2=0$

Incorrect option

Option 4)

$x^{2}+y^{2}-2x-2y-7=0$

Incorrect option

Posted by

Sabhrant Ambastha

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