Question

The line y=3x/4 - 9 is tangent to a circle at (12, 0). The center of the circle is on the line y= 4x/3 - 8. Determine the equation of this circle.

Answer 1

equation is:
\[(x-9)^{2} + (y-4)^{2} = 25\]
you are given slope of tangent line of 3/4
the derivative dy/dx will also be 3/4 at point (12,0)
using implicit differentiation
\[\rightarrow 2(x-h) +2(y-k) \frac{dy}{dx} = 0\]
\[\rightarrow \frac{dy}{dx} = \frac{-(x-h)}{y-k}\]
plug in point and set equal to 3/4 ....solve for k in terms of h
\[k = 16 - \frac{4}{3}h\]
Now we are given line that goes through center (h,k)
\[k = \frac{4}{3}h - 8\]
thus by substitution:
h = 9, k = 4
radius is distance between points (12,0) and (9,4)
r = 5

Answer 2

Fantastic! Thank you so much!

Answer 3

your welcome

Answer 4

The tangent of a circle is perpendicular to its radius, so you can find the equation of the radius by using the slope of the tangent.

Answer 5

And then find the distance between where the tangent line crosses the radius line and the center of this circle, you'll get the radius.

Answer 6

@Montreal_BOY94, c0ckssio, where did you get this question from?
Here is an easier understandable solution:


If I draw a line 'c' that is perpendicular to the line 'b', and goes through the point (12, 0), I get the equation of the line 'c' as:
y = -3x/4+16

This line also intercept with line 'a' at the center of the circle, thus you find the center as:
-3x/4+16 = 4x/3-8
x=9, y=4

DADA