# How to find point of tangency with center of circle and tangent line equation [closed]

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 :)

## closed as off-topic by mlc, HK Lee, Arnaldo, Adam Hughes, The Dead LegendMay 8 '17 at 4:33

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• Yes, there are different methods (not all equally convenient). Have you made an effort to look up something on your own? – mlc May 7 '17 at 21:45

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 $0$: $$10^2-4\cdot5\cdot(10-r^2)=0 \\ \implies r^2=5$$

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.

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: $$(x-3)^2+(y-4)^2=5$$

• I assumed you know formula for distance of a point from a line. – Iti Shree May 7 '17 at 21:39
• The line the circle is tangent to is $y=2x+3$, not $y=x+3$. – George Law May 7 '17 at 21:59
• My mistake, I'll correct that, thanks for telling me @GeorgeLaw – Iti Shree May 7 '17 at 22:02
• The answer I got was $(x-3)^2+(y-4)^2=5$ – mathsisfun.com/data/… – George Law May 7 '17 at 22:27
• Thanks @GeorgeLaw I didn't realized that it'd be $|\frac{-5}{\sqrt5}|$ so many silly mistakes. – Iti Shree May 7 '17 at 22:36

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 $2$