Given that the line y=mx+c is a tangent to the circle
(x-a)^2+(y-b)^2 = r^2
show that:
(1-m^2)r^2=(c-b+ma)^2
Let
L: y=mx+c
C: (x-a)²+(y-b)²=r²
And if L is tangent to C, then
we can solve for the intersections of the line L with circle by substituting "y" in the circle with mx+c, thus
C: (x-a)²+(mx+c-b)²=r²
Expand and solve for quadratic in x to get:
x=[sqrt((m²+1)r²-a²*m²+2am(b-c)-c²+2bc-b²)+m(b-c)+a]/(m²+1)
Since L is a tangent, the two roots must be coincident, hence the terms inside the squareroot sign must vanish, giving
(m²+1)r²-a²*m²+2am(b-c)-c²+2bc-b²=0
Factoring the above expression will give
(m²+1)r²=(c-b+ma)²
[note, this is different from the expected answer you posted. Please check for typo on your part]