Posted by **glysdi** on Friday, September 2, 2011 at 2:14am.

a circle is centered at (-3,4) and in tangent to 5x-4y+12=0.find its equation.

- analytic and solid geometry -
**Reiny**, Friday, September 2, 2011 at 8:19am
equation must be

(x+3)^2 + (y-4)^2 = k , were √k is the radius

2(x+3) + 2(y-4)dy/dx = 0

dy/dx = -(x+3)/(y-4) which is also the slope of the tangent.

But from 5x - 4y + 12 = 0 , the slope of the tangent is 5/4

so (-x-3)/(y-4) = 5/4

5y - 20 = -4x - 12

4x + 5y = 8

4x+5y=8 times 5 ---> 20x + 25y = 40

5x-4y= -12 times 4 --> 20x - 16y = -48

subtract: 41y = 82

y = 2

then in 4x+5y = 8

4x + 10 = 8

x = -1/2 , the point of contact is (-1/2, 2) or (-.5,2)

so back in original:

(-1/2 + 3)^2 + (2-4)^2 = k

k = 41/4 or 10.25

**(x+3)^2 + (y-4)^2 = 41/4**

check:

at (-.5,2) dy/dx = -(-.5+3)/(2-4) = -2.5/-2 = 5/4

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