Question

Find the equations of the tangents to the curve x^2+y^2=25 which pass though the point (-1,7)

Answer 1

using Calculus, let the point of contact of the tangent be P(a,b)
(Did you notice the point (-1,7) is outside the circle ?)

2x + 2y dy/dx = 0
dy/dx = -x/y
so at P, the slope of the tangent is -a/b
but using the old grade 9 way of finding slope,
slope = (b-7)/(a+1)

so b^2 - 7b = -a^2 - a
a^2 + b^2 = 7b-a
25 = 7b-a , since (a,b) lies on x^2 + y^2 = 25
a = 7b - 25

sub that into the circle equation
(7b-25)^2 + b^2 = 25
expanding and simplifying ...
b^2 - 7b + 12 = 0
(b-3)(b-4) = 0

if b=3, then a = -4
if b = 4, then a = 3

for (3,4), slope = -3/4
y = (-3/4)x + b
4 = (-3/4)(3) + b
16 = -9 + 4b
b = 25/4

y = (-3/4)x + 25/4 OR 3x + 4y = 25

leaving the calculation of the second tangent up to you