Question

If the lines $2x-3y-5=0$ and $3x-4y=7$ are diameters of a circle of area $154$ square units, then the equation of the circle is

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

Answer 1

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

Point of intersection of any diagonals give us the center of circle

$2x–3y=5$

$3x–4y=7$

$3\times eq.1–2\times eq.2$

$-9y+8y=15–14$

$-y=1⟹y=-1$

$⟹2x=5+3$

$⟹x=4$

$C=(4,1)$

Given Area $=154=\pi r^2$

$⟹r^2=\frac{154\times 7}{22}=7\times 7$

$⟹r=7$

Equation is

$(x-4{)}^2+(y-1{)}^2=7^2$

$x^2+y^2–8x–2y–47=0$