Question

if the lines $3x-4y-7=0$ and $2x-3y-5=0$ are two diameter of a circle of area $49\pi$ square units the equation of the circle is

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

Answer 1

The correct option is C $x^2+y^2-2x+2y-47=0$

Given that,

Equations of diameter $3x-4y-7=0$ …… (1) and $2x-3y-5=0$…… (2)

Solve equations are

$3x-4y-7=0$ $\times 2$

$2x-3y-5=0$ $\times 3$

$6x-8y-14=0$

$\underset{–}{6x-9y-15=0}onsubtractingthat$

$y+1=0$

$y=-1$

Put the value of in equation (1)

$3x-4y-7=0$

$3x-4(-1)-7=0$

$3x+4-7=0$

$3x-3=0$

$x=1$

Hence , the coordinates of the centre is $(1,-1)$

Also given area of circle= $49\pi$

$\pi r^2=49\pi$

$r^2=49$

$r=7$

Then, we know that the equation of circle is

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

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

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

$x^2+y^2-2x+2y+2=49$

$x^2+y^2-2x+2y=49-2$

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

$x^2+y^2-2x+2y-47=0$

Hence, it is complete solution.

Option $\left(C\right)$ is correct answer.