Given the equation of a circle and a line, find the radius?
The circle (x-1)^2 + (y-3)^2 = r^2 is tangent to the line 5x + 12y = 60
The value of r is…
My book says 19/13, but my answer was a little off
Here’s what I did:
If the line is tangent, then the radius when drawn from the center of the circle to the line is line is perpendicular to the line. Since the equation of the line is 5x + 12y = 60, the slope is -5/12. So, the slope of the radius to the tangent must be 12/5.
So, we have y = -5x/12 + 5, and y = 12x/5 + b. I plugged in (1,3) for the second equation (1,3 being the center), and the equation was
y = 12x/5 – 3/5
I plugged both equations in on my TI-84, and got x and y coordinates of around (1.988, 4.172).
Then I used the distance formula and got 1.53, while 19/13 is 1.46
I know it’s just a little off, but I feel like it shouldn’t be. Did I make any calculation errors?
Northstar: Your answer was the same answer as my book’s, by I didn’t get it….could someone explain it?
When I find the equation of the line through the center and perpendicular to the tangent line, I get
y = 12x/5 + 3/5
(not – 3/5)
I think that will correct your points of intersection and distance calculation.
QED
additional details:
for the formula of the distance from a point to a line in one step, see this link:
http://www.worsleyschool.net/science/files/linepoint/method5.html
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