Find equation of circle tangent to 3x+y+3=0 at (-3, 6)& tangent to x+3y-7=0 Share with your friends Share 1 Rahul Raj answered this Dear student The circle is tangent to 3x+y+3=0eqn of normal through centre isx-3y=kit passes through (-3,6)so x-3y=-21let the centre be (3k-21,k)centre is equidistant from tangents 3x+y+3=0 and x+3y-7=0so3(3k-21)+k+332+12=(3k-21)+3k-712+3210k-6010=6k-281010k-60=6k-2810k-60=±(6k-28)solving we get k=8 and 11/2centre is (3,8) and (-9/2,11/2) radius is 210 and 10/2circles are(x-3)2+(y-8)2=40(x+9/2)2+(y-11/2)2=10/4 Regards 1 View Full Answer Aditi Sharma answered this ugytuytb -2
