Circle equation

Question

Definition of problem:

Write the circle equation which touches the coordinate axis and cross the point $M(2,1).$

I'm confused because I'm used to solve problems with given center but in this problem I cant figure out the formula.

The other problems use the following formula: $$(x-p)^2+(y-q)^2=r^2.$$

So I need a bit help, just to know the way this problem is taking.

By "touches" do you mean that it is tangent to the axes? That is, it intersects each axis at exactly one point? — KSmarts, Jan 21 '15 at 21:27
Can you realize that the points of the axis are $(p,0)$ and $(0,q)$? — DiegoMath, Jan 21 '15 at 21:28
If you solved it, you should probably post your answer for the benefit of others — MPW, Jan 21 '15 at 21:53
How many sets of solution did you get? If there is one, then you must have missed something. If there are two, then why your question’s definition says “write down the circle equation which touches ….”? — Mick, Jan 22 '15 at 11:11

score 1 · Answer 1 · 2015-02-20T09:05:02.817

1

For a circle to tangent both axis in the first quadrant, the center must be at $(r,r)$.

Expressing that the circle is through $(2,1)$:

$$(2-r)^2+(1-r)^2=r^2,$$or $$r^2-6r+5=0=(r-1)(r-5).$$

edited Feb 20 '15 at 09:05

answered Feb 20 '15 at 08:59

score 0 · Answer 2 · answered Feb 20 '15 at 08:47

Given that the circle touches the coordinate axes so ... It should be having coinciding roots at x=0 and y=0 respectively , now second degree eqn for circle is x^2 + y^2 + 2gx +2fy + c= 0 centered at (-g,-f) which is clear and c is any constant here the radius is given by sqrt(g^2+f^2-c), now put the coinciding root condition and you will get g^2=f^2=c(|g| = |f|) , so either g=-f or g=f , out of which g=f only holds true.. , substitute this as well as x=2,y=1 into general equation to make it an equation in one variable either g or f and solve to get coordinates of the center values of g and f and hence the radius finally giving you the answer..

Circle equation

2 Answers2