determine 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)

Question

determine the two tangent line points (x₁,y₁) and (x₂,y₂) on circle (x-h)²+(y-k)²=r² drawn from point (m,n), where (m,n) is outside circle, that is, (m-h)²+(n-k)²>r².

x₁=f₁(r,h,k,m,n)

y₁=f₂(r,h,k,m,n)

x₂=f₃(r,h,k,m,n)

y₂=f₄(r,h,k,m,n)

Answer 1

take a simpler example

circle is x^2 +y^2 =5^2 with center the origin and radius =5

let (m,n) = (0, 10) 5 units above top of the circle

tangent lines are tangent at points with positive y values and +/- x values, on the top half of the circle

tangent lines have slopes = dy/dx = y' where 2 +2yy'= 0, y' =1/y= +/-1/sqr(25-x^2), one positive, one negative. 10 +/-sqr(100-4x) = sqr(25-x^2), solve for x= +/-.93427

points of tangency = (-/+.93427, sqr(25-.93427^2) =about (.93427, 4.9119) and (-.93427, 4.9119)

= about (+/-,9, 4.9)= (m,n) where m= slope of tangent lines, n=y=sqr(25-x^2)

Answer 2

x₁=h+rcos(β+θ)

y₁=k+rsin(β+θ)

β is the angle formed by line though (h,k) and (m,n) measured from x-axis

tanβ=(n-k)/(m-h)

θ is the angle formed by radius line center to (x₁,y₁) measured form (h,k) and (m,n) line

tanθ=[(m-h)²+(n-k)²-r²]^1/2/r

θ is an angle in right triangle with hypotenuse d₂²=(m-h)²+(n-k)² and one leg=r

the second leg d₁²=d₂²-r²

so tanθ=d₁/r

x₁=h+r{(m-h)d₂)(r/d₂)-(n-k)(d₁/d₂)]=h+(r/d₂²)[r(m-h)-d₁(n-k)]

y₁=k+r[(d₁/d₂)(m-h)+(r/d₂)(n-k)/d₂)=k+(r/d₂²)[d₁(m-h)+r(n-k)]

note: if (x₁,y₁) is plugged into formula r²=(x₁-h)(m-h)+(y₁-k)(n-k),,, result is r²=r²

x₂=h+(r/d₂²)[r(m-h)+d₁(n-k)]

y₂=k+(r/d₂²[r(n-k)-d₁(m-h)]

Answer 3

Given is not possible.

If (m, n) is outside of circle then

m = h + r, (m - h) = r. Like wise

n = k + r, (m - k) = r.

Therefore

(m - h)² + (n - k)² = 2r²

The given become

2r² > r² or 2 > 1