Question

The equation of the line tangent to the circle at (x+5)^2+(y-9)^2=289, (-13,-6)

Answer 1

The circle is (x + 5)^2 + (y - 9)^2 = 289 and the point is (- 13, - 6).

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

center of circle is  (h, k ) = (- 5, 9).

The tangent line is perpendecular to the segment joining (- 13, - 6) and(- 5, 9).

Slope of the segment joining (- 13, - 6) and(- 5, 9) = 9 + 6 / - 5 + 13 = 15 / 8.

Because the slopes of perpendicular lines are negative reciprocals, the slope of perpendicular line through (- 13, - 6) is - 8 / 15.

Substitute the values of m = - 8 / 15 and (x₁, y₁) = (- 13, - 6) in equation of tangent line.

y - (- 6) = (- 8 / 15)(x - (- 13))

y +  6 = (- 8 / 15)(x + 13)

y  = (- 8 /15)(x + 13) - 6

y  = (- 8 / 15)x + 13(- 8 / 15) - 6

y  = (- 8 / 15)x + [(- 104 - 90) / 15]

y  = (- 8 / 15)x - (194 / 15).

Therefore, the equation of the line tangent to the given circle is y = (- 8 / 15)x - (194 / 15).