What is the tangent line equation that touches the circle x2 plus 10x plus y2 -2y -39 equals 0 at the point 3 2 0n the Cartesian plane?

Answer 1

A circle with centre (x0, y0) and radius r has the formula:

(x - x0)² + (y - y0)² = r²

Re-arranging x² + 10x + y² - 2y - 39 = 0 into this form using completing the square for x and y gives:

(x + 10/2)² - (10/2)² + (y - 2/2)² - (2/2)² - 39 = 0

→ (x + 5)² - 25 + (y - 1)² - 1 - 39 = 0

→ (x + 5)² + (y - 1)² = 65

→ the circle has centre (-5, 1)

A tangent is perpendicular to a radius.

Its slope m' is the negative reciprocal of the slope m of the radius:

m' = -1/m

The radius to the point (3, 2) has slope:

m = change_in_y/change_in_x

→ m = (2 - 1)/(3 - -5) = 1/8

Thus the slope of the tangent is:

m' = -1/m = -1/(1/8) = -8

The equation of a line with slope m and through a point (X, Y) is given by:

y - Y = m(x - X)

So the equation for the tangent at point (3, 2) which has a slope of -8 is:

y - 2 = -8(x - 3)

→ y - 2 = -8x + 24

→ y + 8x = 26