Question

If the lines $3x-4y-7=0$ and $2s-3y-5=0$ 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

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

Given: Area of circle=$49\pi sq.units$

$\therefore$ Area of circle=$\pi r^2=49\pi sq.units$

where, $r$ is the radius of the circle

$\pi r^2=49\pi$

$r^2=49$

$r=7units$

Given eqs of lines are, $3x-4y-7=0$ and $2x-3y-5=0$

Solving the equations, we get $x=1,y=-1$

The points of intersection of the lines is the center of the circle i.e., $(1,-1)$

We have the eqn for the circle with center $(a,b)$ and radius $r$,

$(x-a{)}^2+(y-b{)}^2=(r{)}^2$

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

$(x^2+1-2x)+(y^2+1+2y)=49$

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

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

$\therefore$ The equation of the given circle is,

$x^2+y^2-2x+2y-47=0$