3
$\begingroup$

Find the equation of a circle that intersects the $y$-axis at the origin and at the point $(0,6)$ and also touches the $x$ axis.

Okay, so I wasn't sure how to do this so I looked at the answer at the back of the book and decided to graph it. The book says the answer is $x^2+y^2-6y=0$ and when graphed it shows that the line joining $(0, 0)$ and $(0, 6)$ passes through the center of the circle and is a diameter. Knowing this made solving the question very easy.

So what I'm kinda confused about is how you would have known, just from the information given in the question or by other methods, that the line joining $(0, 0)$ and $(0, 6)$ was a diameter.

Any other solutions/hints would be much appreciated.

$\endgroup$
1
$\begingroup$

Notice, since the circle intersects y-axis at $(0, 0)$ & $(0, 6)$ hence the line through theses points acts as a chord &

since circle touches the x-axis hence it acts as a tangent. but y-axis (chord) & x-axis (tangent) are normal to each other. Hence this is possible if and only if the line through $(0, 0)$ & $(0, 6)$ must be coincident with the diameter with the circle i.e. acts as a diameter i.e. circle touches x-axis at the origin $(0, 0)$.

Hence $(0, 0)$ & $(0, 6)$ are end points of diameter & its center is the mid point $\left(\frac{0+0}{2}, \frac{0+6}{2}\right)\equiv(0, 3)$ on the y-axis. In this case the radius of circle will be $3$ Hence,

equation of the circle with center $(0, 3)$ & a radius $3$ is given as $$(x-0)^2+(y-3)^2=3^2$$ $$x^2+y^2-6y+9=9$$ $$x^2+y^2-6y=0$$

$\endgroup$
1
$\begingroup$

What do you mean by "touches" the x-axis? Is tangent to the x-axis? Then since it also crosses the y-axis at the origin, the circle must be tangent to the x-axis at (0, 0) and the y-axis is a diameter. Since it also goes through (0, 6) the circle has center (0, 3) and radius 3.

$\endgroup$
1
$\begingroup$

If it is not a diameter, it does not touch but cut the x-axis. Circle centered at $(0, 3)$ with radius 3 is

$$(x-0)^2+(y-3)^2=3^2$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.