Question

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

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

Answer 1

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

R.E.F image

Intersection of two diametrical lines will be center of circle

$\therefore 2x-3y=5$ & $3x-4y=7$

$3(2x-3y=5)\Rightarrow 6x-9y=15$

$2(3x-4y=7)$ $6x-8y=14$

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$-y=1\Rightarrow y=-1$

$\therefore 2x-3x(-1)=5\Rightarrow x=1$

center $(1,-1)$ Radius $\Rightarrow \pi r^2=154sq$

$\frac{22}{7}\times r^2=154\Rightarrow r^2=44\Rightarrow r=7unit$

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

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