Question

If the lines $2x-3y=5$ and $3x-4y=7$ are two diameters of a circle of radius $7$, then the equation of the circle is

• A. $x^2+y^2+2x-4y-47=0$
• B. $x^2+y^2=49$
• C. $x^2+y^2-2x+2y-47=0$
• D. $x^2+y^2=17$

Answer 1

Answer: (C)

Let $\left(x_0,y_0\right)$ be the center of the circle
Intersection of any $2$ diameter gives us the center.
$x=\frac{5+3y}{2}$ substituting this in
$3x-4y=7$
$3\frac{5+3y}{2}-4y=7$
$\Rightarrow y+1=0$
$\Rightarrow y=-1$
$x_0=\frac{5+3(-1)}{2}=1$
Equation of circle will be
${\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2=r^2$
$(x-1{)}^2+(y+1{)}^2=7^2$
$\Rightarrow x^2+y^2-2x+2y-47=0$