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Linear equation passing through a point outside the circle and tangent to the circle

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Find the equation of the straight line tangent to the circle passing through the point P(x1, y1) outside the circle.

The equation of a straight line passing through point P and having slope m is:

y = m*x+y1-m*x1

The equation of a circle with center point O(a, b) and radius r is:

(x-a)^2 + (y-b)^2 = r^2

Since the straight line must meet the circle at a point, substitute the equation of the straight line into the equation of the circle and solve.

(x-a)^2 + (m*x+y1-m*x1-b)^2 = r^2

If c = y1-m*x1-b ,

x^2-2*a*x+a^2+m^2*x^2+2*m*c*x+c^2-r^2 = 0

(m^2+1)*x^2 - 2*(a-m*c)*x + a^2+c^2-r^2 = 0
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Since it must have one real root, the discriminant must be 0.

4*(a-m*c)^2 - 4*(m^2+1)*(a^2+c^2-r^2) = 0

It is expressed as a quadratic equation for m.

(a-m*c)^2-(m^2+1)*(a^2+c^2-r^2) = 0

a^2-2*a*m*c+m^2*c^2-m^2*a^2-m^2*c^2+m^2*r^2-a^2-c^2+r^2 = 0

-2*a*m*c-m^2*a^2+m^2*r^2-c^2+r^2 = 0

-2*a*m*(y1-m*x1-b)-m^2*a^2+m^2*r^2-(y1-m*x1-b)^2+r^2 = 0

-2*a*y1*m+2*a*x1*m^2+2*a*b*m-m^2*a^2+m^2*r^2-(y1-m*x1-b)^2+r^2 = 0

(2*a*x1-a^2+r^2)*m^2-2(a*y1-a*b)*m-(y1-m*x1-b)^2+r^2 = 0

(2*a*x1-a^2+r^2-x1^2)*m^2-2(a*y1-a*b-x1*y1+x1*b)*m-y1^2-b^2+2*b*y1+r^2 = 0

Here,

A = (2*a*x1-a^2+r^2-x1^2)
B = -2*(a*y1-a*b-x1*y1+x1*b)
C = -y1^2-b^2+2*b*y1+r^2

, and obtain the slopes m1, m2 and the y-intercepts b1, b2, respectively, using the quadratic formula.

m1 = -(1/(2*A))*(B + sqr(B^2-4*A*C))

b1 = y1-m1*x1

m2= -(1/(2*A))*(B - sqr(B^2-4*A*C))

b2 = y1-m2*x1

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