Question

The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle having area as 154 sq. units. Then the equation of the circle is

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

Answer 1

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

Center is the point of intersection of diagonals.

$\Rightarrow 2x-3y=5$

$\Rightarrow 3x-4y=7$

Solving simultaneous equation, $x=1,y=-1$

Center $=(1,-1)$

For radius $r$ ( let )

$\Rightarrow r=\sqrt{\frac{154}{\pi }}$ $\Rightarrow \sqrt{\frac{154}{22}\times 7}=7$

Equation -

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

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

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

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

Hence, the answer is $x^2+y^2-2x+2y=47.$