Finding the points where a circle intersects an axis

Question

A circle has the equation:

x²+y²+4x-2y-11 = 0

What would be the coordinates of the points where the circle intersects with the y-axis and how would you calculate it?

Answer 1

$x^2+y^2+4x-2y-11=0$

The points of intersection with the $y$ axis are when $x=0$. Thus, plug in $x=0$ and solve.

$$0^2+y^2+4(0)-2y-11=0$$ $$y^2-2y-11=0$$ Use the quadratic formula: $\large\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ where $a=1, b=-2, c=-11$

$$\frac{-(-2)\pm \sqrt{4-(-44)}}{2}$$ $$\frac{2\pm \sqrt{48}}{2}$$

$\sqrt{48}=\sqrt{8\cdot 6}=\sqrt{4\cdot 2\cdot 3\cdot 2}=4\sqrt{3}$

$$\frac{2\pm 4\sqrt{3}}{2}=1\pm 2\sqrt{3}$$

So, your coordinates for intersection with the $y$-axis are: $(0,1+ 2\sqrt{3})$ and $(0,1- 2\sqrt{3})$

Answer 2

Method1:

Given equation is $(x+2)^2+(y-1)^2=(4)^2$.

Now y-axis corresponds to $x=0$, let it intersect y-axis at $(0,k)$,So , $$(y-1)^2=16-4=12\\ y-1=\pm2\sqrt3\\ y=1\pm2\sqrt3$$

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Method2:

Given circle: $$x^2+y^2+4x-2y-11 = 0$$ Now y-axis corresponds to $x=0$, so let it intersect y-axis at $(0,k)$: $$k^2-2k-11=0$$ Using quadratic formula: $$k=\frac{2\pm\sqrt{2^2-(-2)(11)}}{2(1)}=1+2\sqrt3,1-2\sqrt3$$