Find the coordinates of the point of contact of these two circles?

Question

(x-4)^2 + (y-3)^2 = 40

x^2 + y^2 - 2x + 12y + 27 = 0

Ive done about 3 pages of working out and I keep getting it wrong ><

Halp?

Answer 1

The first circle is centered at (4,3) and has a radius of 2SQRT(10)

Now the second circle:

(x^2 - 2x + 1) - 1 + (y^2 + 12y + 36) - 36 + 27 = 0

(x - 1)^2 + (y + 6)^2 =10

So the second circle is centered at (1,-6) and has a radius of SQRT(10)

Distance between the centers of the circles is:

D = SQRT[(1 - 4)^2 + (-6 - 3)^2] = SQRT(90) = 3SQRT(10)

So the distance between the centers is 3SQRT(10) which means that the circles intersect at one point along the line that joins the centers since the sum of their radii is 3SQRT(10).

There are different ways you can go about this but let us try a line through the centers. We know that the intersection point will be a distance SQRT(10) along this line from the center of the second circle.

First we need the slope of this line:

Slope of line between centers = m = (-6 - 3) / (1 - 4) = 3

We can now use this slope and the radius of the circle to find the intersection point. We want to add offsets from (1,-6) to give the point of intersection. Since this is just a triangle we can use sines and cosines where the angle is:

tan(Angle) = slope = 3

I sometimes use triangles for this so the sides are 3, 1 and SQRT(10)

sin(Angle) = 3/SQRT(10)

cos(Angle) = 1/SQRT(10)

x(intersection) = X = x(center) + radius*cos(Angle)

y(intersection) = Y = y(center + radius*sin(Angle)

X = 1 + SQRT(10)*[1/SQRT(10)] = 2

Y = -6 + SQRT(10)*[3/SQRT(10)] = -3

So the point of intersection is (2,-3)

Check:

First circle: (2 - 4)^2 + (-3 - 3)^2 = 2^2 + 6^2 = 40

Second circle: 2^2 + (-3)^2 - 2(2) + 12(-3) + 27 = 4 + 9 - 4 - 36 + 27 = 0

So the point just found does lie on both circles and is therefore the point of intersection.

Answer 2

You have two eqns in two unknowns

x^2 - 8x + 16 + y^2 - 6y + 9 = 40 and

x^2 - 2x        + y^2 + 12y + 27 = 0

Rewriting:

x^2 - 8x + y^2 - 6y = 15

x^2 - 2x + y^2 + 12y = -27

Subtracting the second from the first:

6x + 18y = 42 and dividing 6

x + 3y = 7 or y = -x/3 + 7/3 a straight line

Where does this straight line intersect each circle; there should be one point where this occurs

You do this by putting in the value of y into each eqn, finding the value for x, then finding the corresponding value(s) of y.  This you can do and will be the answer.