Prove the two tangent line points (x1,y1) and (x2,y2) on circle (x-h)^2+(y-k)^2=r^2 drawn from point (m,n) satisfy equations r^2=(x1-h)(m-h)+(y1-k)(n-k) and r^2=(x2-h)(m-h)+(y2-k)(n-k)

Question

any circle (x-h)²+(y-k)²=r²

a point outside the circle (m,n),,,,that is (m-h)²+(n-k)²>r²

has two lines through point tangent to the circle, on the circle,

at points (x₁ ,y₁) and (x₂,y₂)

show that these two points satisfy the equations

r²=(x₁-h)(m-h)+(y₁-k)(n-k)

and

r²=( x₂-h)(m-h)+(y₂-k)(n-k)

Answer 1

for (x₁,y₁)

slope of radius line m_r=(y₁-k)/(x₁-h)

slope of tangent line to circle m_t=(n-y₁)/(m-x₁)

m_r=-(1/m_t) is consequence of orthogonality

(y₁-k)/(x₁-h)=(x₁-m)/(n-y₁)

-y₁²+(k+n)y₁-kn=x₁²-(h+m)x₁+hm

know x₁²+y₁²=r²+2hx₁-h²+2ky₁-k²

r²=(k+n)y₁-kn+(h+m)x₁-hm-2hx₁+h²-2ky₁+k²

r²=(n-k)y₁-kn+k²+(m-h)x₁-hm+h²

r²=(n-k)y₁-k(n-k)+(m-h)x₁-h(m-h)

r²=(x₁-h)(m-h)+(y₁-k)(n-k)

It is a curious result,,a little like Bezout's Identity 1=xA+yB

same algebra used for (x₂,y₂)

Answer 2

If (x₁, y₁) lie on circle, then (x₁ - h)² + (y₁ - k)² = r².

To satisfy the condition (x₁ - h)(m - h) + (y₁ - k)(n - k) = r²

(m - h) = (x₁ - h) and (n - k) = (y₁ - k). Resulting in

(m, n) = (x₁, y₁) and (m, n) is not outside circle.