For what values of $a$ will $y=ax$ be a tangent to $x^2+y^2+20x-10y+100=0$

Question

For what values of $a$ will $y=ax$ be a tangent to $x^2+y^2+20x-10y+100=0$

I tried to solve this question by differentiating and making it equal to $0$ and solving for $x$ and i got $-10$ as an answer but when i graph the equation, it doesn't touch the circle. I don't know where i gone wrong. Can anyone help me with this question.

Thanks.

Answer 1

Since $x^2+y^2+20x-10y+100=0$ can be written as $$(x+10)^2+(y-5)^2=5^2,$$ we know that the center of the circle is $(-10,5)$ with radius $5$.

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Hence, it is easy to see that $a=\color{red}{0}$ is a solution where $(-10,0)$ is the tangent point.

Also, noting that $\tan\alpha=5/10=1/2$, we have $$a=\tan\theta=\tan(\pi-2\alpha)=-\tan(2\alpha)=-\frac{2\tan\alpha}{1-\tan^2\alpha}=\color{red}{-\frac 43}.$$

Answer 2

A purely algebraic alternative.

If the line $y=ax$ is tangent to the circle $x^2+y^2+20x-10y+100=0$, then they should only have 1 point in common.

Substitution of $y=ax$ into the equation of the circle gives you a quadratic equation in $x$ with parameter $a$. This equation has exactly 1 solution if the discriminant is 0, so: $$x^2+(ax)^2+20x-10ax+100=0 \Leftrightarrow (1+a^2)x^2+(20-10a)x+100=0$$ The discriminant is $(20-10a)^2-400(1+a^2)$, so: $$(20-10a)^2-400(1+a^2) = 0 \Leftrightarrow -100a(3a+4) = 0 \Leftrightarrow \color{blue}{a = 0} \vee \color{blue}{a = -\frac{4}{3}}$$ So we find the same as in the answer with the more geometric approach.