Question

Find the equation of circle, if the lines $2x-3y=5$and $3x-4y=7$are diameters of a circle of area 154 sq units.

Answer 1

Since , $2x-3y=5$ and $3x-4y=7$ are diameters of circle.

On solving these two equations. we get

$x=1$and $y=-1$

We know that point of intersection of diameters of circle is the centre of circle.

$\therefore$Centre of circle is (1,-1)

Also given , area of circle =154

$\Rightarrow \pi r^2=154\Rightarrow r^2=\frac{154}{22}\times 7$

$\therefore r=7$

$\therefore$ Required equation of circle is

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

$\Rightarrow (x-1{)}^2+(y+1{)}^2=(7{)}^2$

$\Rightarrow x^2-2x+1+y^2+2y+1=49$

$\Rightarrow x^2+y^2-2x+2y=47$