Question

The lines $2x-3y-5=0$ and $3x-4y=7$ are diameters of a circle of area $154squnits$, then the equation if the circle is :

• A. $x^2+y^2+2x-2y-62=0$
• B. $x^2+y^2+2x-2y-47=0$
• C. $x^2+y^2-2x+2y-62=0$
• D. $x^2-y^2-2x+2y-62=0$

Answer 1

The correct option is A $x^2+y^2+2x-2y-62=0$

$\pi r^2=154$

$r=7$

For centre on slowing equation we get,

$2x-3y=5$ …………$\left(1\right)$

$3x-4y=7$ ………….$\left(2\right)$

Multiply $3$ to equation $\left(1\right)$ and Multiply $2$ to equation $\left(2\right)$

$6x-9y=15$

$6x-8y=14$

_____________

$-y=1$

$y=-1$

$x=1$

So centre is $(1,-1)$

Equation of the circle $(x-1{)}^2+(y+1{)}^2=7^2$

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

$\Rightarrow x^2=y^2-2x+2y-47=0$.