Question

From point p tangents equal lenghts find equation of circle through p which touches x+y=5 at point (6,-1)

Answer 1

Center of circle lies on the
line x-y-1=0.Let C(a+1,a) be center of the circle.Length of perpendicular from C to tangent 3x+4y-35=0 is given by

CM=(3(a+1)+4a-35)/

5=(7a-32)/5
Radius of the circle is

given by

R=√((a+1-5)^2+(a-5)^2)
Using condition of tangency

R=CM we get
√((a-4)^2+(a-5)^2)=(7a-32)/5

25(2a^2-18a+41)=49a^2-448a+1024
50a^2-450a+1025=49a^2-448a+1024
a^2-2a+1=0
a=1.