ambuj tiwari
Last Activity: 7 Years ago
Let eq of circle is x^+y^+2gx+2fy+c=0. and according to question point (2,5) is a point on circle on which tangent drawn so it is satisfied in this equation. after put it we get. 4g+10f+29+c=0. And. Centre (-g,-f) satisfy in given equation x-2y=4. So. 2f-g=4. And the perpendicular distance from 2x-y+1=0. From centre of circle is radius. So. -2g+f+1/root5=root over g^+f^-c. Now here three equation and three variable then solve it and finally answer is 3root5