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
$3x-4y-7 =0 \ldots \left(i\right)$
and $\,$ $2x-3y -5 =0 \ldots.\left(ii\right)$
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
$\pi r^{2}=49\pi$
$\Rightarrow \, \quad r=7$ units
$\therefore $ Equation of required circle is
$\left(x-1\right)^{2}+\left(y+1\right)^{2} =49$
$\Rightarrow \, x^{2}+y^{2}-2x +2y+1+1=49$
$\Rightarrow \, x^{2}+y^{2}-2x +2y-47=0$