Question

Find the equations of the tangent from the origin to the circle
x^2+y^2-5x-5y+10=0

Answer 1

differentiate implicitly ...

2x + 2y dy/dx - 5 - 5dy/dx = 0
dy/dx (2y - 5) = 5 - 2x
dy/dx = (5-2x)/(2y-5)

let the point of contact be (a,b)
slope of tangent = (5-2a)/(2b-5)
but the slope of the tangent from (a,b) to (0,0) = b/a

so (5-2a)/(2b-5) = b/a
5a - 2a^2 = 2b^2 - 5b
2a^2 + 2b^2 = 5a + 5b
2(a^2+b^2) = 5(a+b)
a^2 + b^2 = (5/2)(a+b)

but since (a,b) lies on the curve,
a^2 + b^2 - 5a - 5b + 10 = 0
sub in the above
(5/2)(a+b) - 5a - 5b =- 10
times 2
5(a+b) - 10a - 10b = -20
-5a-5b = -20
a+b = 4
a = 4-b
sub into a^2 + b^2 = (5/2)(a+b)
(4-b)^2 + b^2 = (5/2)(4-b + b)
b^2 - 8b + 16 + b^2 = 10
2b^2 - 8b + 6 = 0
b^2 - 4b + 3 = 0
(b-1)(b-3) = 0
b = 1 or b = 3
then a = 3 or a = 1

There will be 2 points of contact, (1,3) and (3,1)

I will do one of the tangents, you do the other one

1. for the point (1,3) and the origin,
slope of tangent = 3/1 = 3
equation of tangent is y = 3x