analytic geometrycircles
So this was a question that I was solving that got me stuck. Its as follows:
Q. Find equation of diameter of the circle $x^2 + y^2 – 6x + 2y = 0$ which passes through the origin.
Now I have tried the following,
From the equation I found,
$g=(-3)$ and $f=1$ and subsequently found the radius as $r=\sqrt{10}$ and hence diameter as $d=2\sqrt{10}$.
But I cant seem to understand how to progress, help would be greatly appreciated.
I am not sure what is asked exactly in the question but if my guess is true it asks the equation of black circle
Here $B$ denotes the center of the blue circle, $A$ denotes the center of the red circle and $H$ the center of the black circle. We know the coordinates of $B$ and $A$ hence we can find the equation of green line. After that if we solve the equation of the line with the red and the blue circles we can find coordinates of $E$ and $G$. We know the diameter of the big circle is the distance between $E$ and $G$ and the center of the big circle is mid point of $EG$. Therefore we can find the equation of the big circle.
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 $$(x-0)^2+(y-3)^2=3^2$$ $$x^2+y^2-6y+9=9$$ $$x^2+y^2-6y=0$$
Best Answer
Hint: Complete the squares in $x$ and $y$.