What is the equation of the tangent line when it touches the circle x square plus y square -2x -6y plus 5 equals 0 at the point of 3 4?

Answer 1

x² + y² - 2x - 6y + 5 = 0

→ x² - 2x + 1 - 1 + y² - 6y + 9 - 9 + 5 = 0

→ x² - 2x + 1 + y² - 6y + 9 = 9 - 5 + 1

→ (x - 1)² + (y - 3)² = 5

→ Circle has centre (1, 3).

Slope of radius line from centre to point (3, 4) is:

m = (4-3)/(3-1) = 1/2

The tangent is perpendicular to the radius line; therefore its slope m' is such that:

mm' = -1

→ m' = -1/m = -1/(1/2) = -2

Therefore the tangent line that passes through the point (3, 4) with slope -2 is:

y - 4 = -2(x - 3)

→ y = -2x + 6 + 4

→ y + 2x = 10

Answer 2

x^2 + y^2 - 2x - 6y + 5 = 0=> x^2 - 2x + 1 + y^2 - 6y + 9 = - 5 + 1 + 9 = 5

=> centre C = (1, 3)

Then, if T = (3, 4), gradient of CP = (4-3)/(3-1) = 1/2

=> gradient of tangent = -2

=> equation of tangent = y = - 2x + c

and, since T = (3,4) is on the tangent, 4 = -6 + c => c = 10

So, the equation of the tangent is 2x + y = 10