write the equation of the line that is tangent at the given point (x-2)^2+(y-4)^2=289;(-15,4)

Question

Using slope-point formula.

Answer 1

The circle is (x - 2)^2 + (y - 4)^2 = 289 and the point is (- 15, 4).

Equation of tangent line : y - y₁ = m (x - x₁).

center of circle is  (h, k ) = (2, 4).

The tangent line is perpendecular to the segment joining (- 15, 4) and (2, 4).

Slope of the segment joining (- 15, 4) and (2, 4) = (4 - 4)/(2 + 15) = 0/17 = 0.

Because the slopes of perpendicular lines are negative reciprocals, the slope of perpendicular line through (- 15, 4) is also - 1/0.

Substitute the values of m = - 1/0 and (x₁, y₁) = (- 15, 4) in equation of tangent line.

y - 4 = (- 1/0)(x - (- 15))

(y - 4)0 = - 1(x + 15)

- x - 15 = 0

x = - 15.

Therefore, the equation of the line tangent to the given circle is x = - 15.