Question

C(3,2) and tangent line passing through (-1,-3) and (-4,0).Find the equation of a circle?

Answer 1

With only the center of this circle given, all we can do up front is to write out the general equation of a circle with radius r and center (h,k):

\[(x-h)^2 + (y-k)^2 = r^2\rightarrow(x-3)^2+(y-2)^2=r^2\]

Answer 2

To proceed further, we'll have to put to use the fact that one of the tangent lines to this circle passes through (-1,-3) and (-4,0). My reasoning is that one point on the circle has to lie on this tangent line, and that the slope of the tangent line can be expressed in two different ways: (1) using the slope formula and the two given points, and (2) finding the derivative of the equation of the circle (given above), which represents the slope of the tangent line.

Answer 3

(1): The slope of the line thru (-1,-3) and (-4,0) is\[m=\frac{ -3-0 }{ -1-[-4] }=?\]

Answer 4

(2) The slope of the tangent line to the circle is found through implicit differentiation:\[\frac{ d }{ dx }[(x-3)^2+(y-2)^2=r^2]\]

Answer 5

Equating those two slopes produces an equation in x and y.

Finally, we can find the equation of the straight line through the two given points using the point-slope formula. This work also produces an equation in x and y.

Solving these two equations will likely result in our being able to identify the coordinates of the point of tangency. With that info, we can find the radius of the circle.

@onepiece?