Question

If the lines $2x-3y=5$ and $3x-4y=7$ are diameters of a circle of area $154$ square units, then obtain the equation of the circle.

Answer 1

Given: $2x-3y=5$ … (i)

and $3x-4y=7$ … (ii).

Solving equation (i) and (ii) we get $(1,-1)$.

So, the centre of circle is: $(1,-1)$.

Let $r$ is the radius of circle.

So, area of circle $=\pi r^2=154$

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

$\Rightarrow r=7$

Equation of circle having centre $(x_1,y_1)$ and radius $=a$ is given by

$(x-x_1{)}^2+(y-y_1{)}^2=a^2$.

$\Rightarrow$ Equation of required circle is: $(x-1{)}^2+(y+1{)}^2=(7{)}^2$

$[\because Centreis(1,-1)andradiusr=7]$

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

Hence, equation of required circle is $x^2+y^2-2x+2y=47$