Thread: Another question on Circle

I have no idea to deal with this question. Can anyone give me some hints?

A circle passes through the point A(2, -1) and is tangent to the line y= x - 1. Its centre is on the line 2x + y = 0. Find the equation of the circle.

The red line is the tangent line.
The green line is where the center lies on.
A circle dot represents a point which lies on the circle.
A rhombus dot represents the center of circle.


This is what I did.
1)Drew the red line and green line and circle dot (given to us).
2)Pick a point (x,y) on the green line and find locus that is equidistanct from the intersection point (1/3,-2/3) and the dot (2,-1).
3)The locus was a line (not show), it intersected the green line at (1,-2)-the rhombus dot, and that was the center of circle.
4)Now we have a center (1,-2) and a point on it (2,-1) we can find the distance, sqrt(2), and find the equation of circle.
(x-1)^2+(y+2)^2=2

Hello, ling_c_0202!

I can give you some hints . . .

A circle passes through the point A(2, -1) and is tangent to the line L: y = x - 1.
Its centre is on the line M: 2x + y = 0.
Find the equation of the circle.

It certainly helps to make a sketch.
Of course, we want the center and radius of the circle.


The center is on line M: .y = -2x
So the center C has coordinates: (p, -2p)

Then the distance from C to A is the radius.
. . sqrt{9p - 2)² + (-2p + 1)²} .= .r [1]

You need to find the perpendicular distance from C to line L.
. . That too is the radius r.
When you find that, you'll have another equation in p and r. [2]
. . Solve the system of equations.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I have found two solutions.
. . . . . . . . . . . . _
. . C(1, -2), r = √2
. . . . . . . . . . . . . . ._
. . C(9, -18), r = 13√2

And they both check out. .Anyone care to double-check them?

Originally Posted by Soroban

[size=3]
. . sqrt{9p - 2)² + (-2p + 1)²} .= .r [1]

Thank you very much! I have got the answers!