SOLUTION: Given a circle: (x-4)^2 + (y+5)^2 = 5, find the equation of the line (in slope-intercept form) that is tangent to this circle at the point (3,-7). Hint: use the fact that a tangent

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The problem tells you that (3,-7) is a point on the circle, but is it really?
Check this by filling in 3 for x and -7 for y
(x-4)^2 + (y+5)^2 = 5
(3-4)^2 + (-7+5)^2 = 5
(-1)^2 + (-2)^2 = 5
1 + 4 = 5
So (3,-7) really is on the circle
Where is the center of the circle?
The equation is in the form
(x - a)^2 + (y - b)^2 = r^2
where (a,b) is the center, and r is the length of the radius
So the center is (4,-5) and r^2 = 5, so r =
Now you have the endpoints of the line, so use

where


and


The key is to get this equation in the form
then the slope of any line perpendicular to it will have the
slope


multiply both sides by (x - 3)


The slope m = 2, so the slope of a line perpendicular to this line
will have slope
This is the slope of the tangent line at (3,-7), so you can write


multiply both sides by 2

multiply both sides by x-3



Does this line go through (3,-7) ? Check it




OK