If the lines 3x-4y-7=0 and 2x-3y-5=0 are two diameters of a circle of area 49π square units, then what is the equation of the circle?

Solution :

Area = 49π

π$r^2$ = 49π

r = 7

Now find the coordinates of center of circle by solving the given two equations of diameter.

By solving the above equation through elimination method we get,

x = 1 and y =-1

which are the coordinates of center of circle.

Now the general equation of circle is $(x-a)^2$  + $(y-b)^2$ = $r^2$

$(x-1)^2$  + $(y+1)^2$ = $7^2$

$(x-1)^2$  + $(y+1)^2$ = 49

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