How to find point of tangency with center of circle and tangent line equation

Question

Find the equation of the circle with center $(3,4)$ that is tangent to the line whose equation is $y = 2x + 3$.

I know you can use the distance between a point and a line formula but according to my math teacher, there's another way to solve it by finding the point of tangency first. Any help would be appreciated :)

Answer 1

The equation of circle will be $$(x-3)^2+(y-4)^2=a^2$$, where $a$ is radius of the circle.

Now to find the radius of circle find distance of center (3,4) from tangent, $y=2x+3$ which would be : $$ | \frac{-5}{\sqrt5}|$$ that is $\sqrt5$

so your equation is: $$(x-3)^2+(y-4)^2=5$$

Answer 2

Let the point of tangency be $(a,b)$. The line joining the centre of the circle to this point is parallel to the vector $$\mathbf v\ =\ \begin{pmatrix}a-3 \\ b-4\end{pmatrix}$$ The line $y=2x+3$ is parallel to the vector $$\mathbf w\ =\ \begin{pmatrix}1 \\ 2\end{pmatrix}$$ (it has gradient $2$). The two vectors are orthogonal, so their dot product is zero: $$\begin{align*}\mathbf v\cdot\mathbf w\ &=\ 0 \\ \implies\quad(a-3)\cdot1+(b-4)\cdot2\ &=\ 0 \\ \implies\quad a+2b\ &=\ 11\ \ldots\ \fbox1\end{align*}$$ As $(a,b)$ lies on the line $y=2x+3$, i.e. $2x-y=-3$ we have $$2a-b\ =\ -3\ \ldots\ \fbox2$$ Now solve $\fbox1$ and $\fbox2$ to find the point of tangency; the distance between this point and $(3,4)$ will be the radius of the circle.

Answer 3

Let the equation of the circle be $$(x-3)^2+(y-4)^2=r^2$$

Substituting $y=2x+3$, $$(x-3)^2+(2x+3-4)^2=r^2 \\ (x-3)^2+(2x-1)^2=r^2 \\ x^2-6x+9+4x^2-4x+1=r^2 \\ 5x^2-10x+10-r^2=0$$ If the circle is tangent to the line, the last equation should yield a unique solution for $x$, i.e. the discriminant is $0$: $$10^2-4\cdot5\cdot(10-r^2)=0 \\ \implies r^2=5$$

Answer 4

Hint: Formula for equation of tangent:

$$(y-4)=m(x-3) \pm r\sqrt{1+m^2}$$ correlate it with $$y=2x+3$$

Your $m$ is $2$