Question

Let S1=x2+y2-4x-8y+4=0 and S2 be its image in the line y=x, find the equation of circle touching y=x at (1,1) and its radical axes with S2 passes through the centre S1.​

Answer 1

Solution :-

S= ( x+ 7/2)^2 + (y-11/2)^2 = 81/2

(Please check Attachment file )

I hope it's help you :)

Answer 2

Answer:

Use all the informations given and follow the step line wise.

First find the S2 circle eqn.

For that first find the image of center of circle S1 along y=x.

U can do this by solving the eqns y=x and the eqn. woyh slope -1 and passing through center of S1.

then asssume radical axis eqn be y = mx +c and pass it through center of S1. as given.

One more information is given:

apply perpendicular distance formulae for eqn. given from center of the circle S1.

Club all the eqns. and get the suitable measure and concclude the final result.