Find point of intersection of tangent and circle given 3 points

Question

I'm told that a circle intersects the y-axis at $(0,0)$ and $(0, -4)$.

I have a tangent that starts at $(0, -6)$. I want to find the point of intersection.

I am taking the midpoint of the circle to be $(0, -2)$ and the radius to be $2$ from the points above.

The equation of the circle then is:

${(x - 0)^2 + (y + 2)^2 = 4}$

$\implies$ ${x^2 + y^2 + 4y = 0}$

I am taking the equation of the line to be ${y = -6}$.

If I substitute ${y = -6}$ into ${x^2 + y^2 + 4y = 0}$ I get

${x^2 + 36 - 24 = 0}$

${x^2 = -12}$

I think I have gone wrong somewhere.

Answer 1

Your assumption that the tangent at (0,-6) is y=-6 is incorrect as this line doesn't touch the circle and therefore you have no point on the real plane that satisfies both the circle and the line.

You will have to find the equation of tangent using its property that the perpendicular distance of line from centre is equal to the radius.

Answer 2

I am taking the equation of the line to be ${y = -6}$.

If I substitute ${y = -6}$ into ${x^2 + y^2 + 4y = 0}$ I get

${x^2 + 36 - 24 = 0}$

${x^2 = -12}$

I think I have gone wrong somewhere.

You have not done anything wrong, the reason why this has happens is because the line $y=-6$ does not intersect the circle. This is why you get ${x^2 = -12}$ which has no real roots. Which is exactly as it should be.

But to get any further with this question we need some more information from you:

I have a tangent that starts at $(0,−6)$.

A tangent to what?

I am taking the midpoint of the circle to be $(0,−2)$.

By midpoint I presume you mean circle centre. But what makes you think this is at $(0,−2)$?

I want to find the point of intersection.

Point of intersection of what with what? (I'm assuming one of those is the circle)

I am taking the equation of the line to be $y=−6$.

Why? Equation of what line? The tangent to the circle?

Answer 3

see this, it shows that your answer is not correct