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

Explanation

Given line are `2x-3y-5=0` ….(i)
and `3x-4y-7=0`
From Eqs. (i) and (ii), `x/(21-20)=y/(-15+14)=1/(-8+9)`
`rArr x/1=y/(-1)=1/(+1)`
`rArr xpm1,y=-1`
Since the interception point of these lines will be coordinates of the circle i.e., coordinates of teh circle as (1,-1). ltbr. Let the radius of the circle is r.
Then `pir^(2)=154`
`rArr22/7xxr^(2)=154`
`rArr r^(2)=(154xx7)/22`
`rArr r^(2)=(14xx7)/2rArrr^(2)=49`
So, the equation of the circle is
`(x-1)^(2)+(y+1)^(2)=49`
`rArr x^(2)-2x+1+y^(2)+2y+1=49`
`rArr x^(2)+y^(2)-2x+2y=47`