find the equation of the circle(s) tangent to the line 3x-4y-16=0 and containing the points (1,8) and (-2, -1)

Question

there will be two equations of the circle.

Answer 1

Here's how to solve the problem:

First find the center of the circle (call it (x1,y1).
1. Write an expression for the distance of (x1,y1) from the first point, (1,8): (x1-1)^2 + (y1-8)^2

2. Write an expression for the distance of (x1,y1) from the second point, (-2,-1). (You should know how to do this, especially after seeing the example above.)

3. Write an expression for the distance of (x1,y1) from the given line. This is a little trickier. There are probably various ways of approaching it. Here's one:

Rewrite the equation in slope-intercept form: y = (3/4)x +4
This tells us that the slope of the line is 3/4. So the slope of a line perpendicular to this line is -4/3. (The product of the two slopes has to be -1.)

The distance from (x1,y1) to the original line is measured along the direction perpendicular to that line. So it will be measured along a line with a slope of -4/3. The general equation for such a line is
y = -(4/3)x + b, where b is the y-intercept of the line.

Now we're trying to find an expression for the distance from (x1,y1) to the original line. To do this, we have to find where the perpendicular line intersects the original line. So we need to find the solution of the two equations,
y = (3/4)x +4 and y = -(4/3)x + b.
This gives us two expressions for y, so we can set them equal:
(3/4)x +4 = -(4/3)x + b
Solving this, x = (12/25)(b-4) = (12b-48)/25
And y = 9/25(b-4) + 4 = (9b +64)/25
So the point of intersection of the given line and the perpendicular line is ((12b-48)/25,(9b +64)/25).

The distance from (x1,y1) to this point (and to the given line) is
sqrt((x1-(12b-48/25))^2 + (y1-(9b+64)/25)^2).

How do we find b and x1 and y1? We need to equate the three expressions for distance (this one, and the two found in items 1 and 2 above). Solving these equations will give us the center of the circle and the equation for the perpendicular to the given line.

Next find the radius of the circle:
Determine the distance from the center to either of the two given points (using the Pythagorean theorem ... that is, using the distance formula). Call this distance D.

Finally, write the equation of the circle:
(x-x1)^2 + (y-y1)^2 = D^2

Answer 2

Hint: think about finding the center of the circle.
You know that it would be equal distance from both the 2 points, and from the line thats perpendicular to it!

OKay, i used y=(-1/3)x+(20/6), y=(-4/3)x+b, c=(3/4)d-4, c=(-4/3)d+b, and sqrt((x-1)^2+(y-8)^2) = sqrt((x-d)^2+(y-c)^2).

And I got a center for one of the circles (1,3).

Answer 3

im not finding shlt!