Find the equation of a circle.

Question

If a circle which center is on the straight line $x-2y+3=0$ cuts both the x-axis and the y-axis, what is the equation of the circle?

ANSWER: $x^2+y^2-6x-6y+9=0$ This is rather a straightforward question but I don't know how to start...some hints would be very much appreciated.

Does 'cuts' mean 'touches'? – lab bhattacharjee Oct 13 '13 at 11:07 — lab bhattacharjee, Oct 13 '13 at 11:07
@labbhattacharjee I assumed so too. – Sophia Oct 13 '13 at 11:19 — Sophia, Oct 13 '13 at 11:19

score 1 · Answer 1 · answered Oct 13 '13 at 11:10

1

HINT:

Let $\alpha,\beta$ be the center of the circle with radius $=r$

$\implies \alpha-2\beta=3\iff \alpha=2\beta-3$

Assuming 'cuts' means 'touches', the axes are tangents of the circle.

So, the perpendicular distance(s) of each axis from the center$(2\beta-3,\beta)$ equals to the radius$(=r)$

Do you know the equation of the axes and

how to calculate the perpendicular distance of a straight line from a given point

answered Oct 13 '13 at 11:10

lab bhattacharjee

@Sophia, We shall have $|\beta|=r=|\alpha|=|2\beta-3|$ Squaring we get $\beta^2=(2\beta-3)^2\implies \beta=?$ – lab bhattacharjee Oct 13 '13 at 11:18
Okay, I'll try to solve it! ;D – Sophia Oct 13 '13 at 11:19
@Sophia, there are two circles, right? I have used one of the two definitions of the tangent of a circle – lab bhattacharjee Oct 13 '13 at 11:27

score 0 · Answer 2 · answered Oct 13 '13 at 11:07

0

Hint: Start with $\left(x-2a+3\right)^{2}+\left(x-a\right)^{2}=r^{2}$

answered Oct 13 '13 at 11:07

drhab

2 Answers2