Question

If the lines $3x-4y-7=0$ and $2x-3y-5=0$ are two diameters 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

Answer: (C)

The given equations of diameters are
$3x-4y-7=0\dots \left(i\right)$
and $2x-3y-5=0\dots .\left(ii\right)$
On solving Eqs. (i) and (ii), we get
x = 1 and y = -1
$\therefore$ Centre of circle is (1,-1).
Let r be the radius of circle, then
$\pi r^2=49\pi$
$\Rightarrow r=7$ units
$\therefore$ Equation of required circle is
${\left(x-1\right)}^2+{\left(y+1\right)}^2=49$
$\Rightarrow x^2+y^2-2x+2y+1+1=49$
$\Rightarrow x^2+y^2-2x+2y-47=0$