# What is the equation of a circle that has its center at the origin and is tangent to the line with equation 3x-4y=10?

May 5, 2018

The circle's equation would be ${x}^{2} + {y}^{2} = 4$

#### Explanation:

We know that a circle's equation centered at the origin would look something like ${x}^{2} + {y}^{2} = {r}^{2}$, where $r$ is the radius. We want to find $r$ so that we can have the circle's full equation.

We can make a system of equations:

$\left\{\begin{matrix}3 x - 4 y = 10 \\ {x}^{2} + {y}^{2} = {r}^{2}\end{matrix}\right.$

We can solve for $y$ in the first equation:

$3 x - 4 y = 10$

$- 4 y = 10 - 3 x$

$4 y = 3 x - 10$

$y = \frac{3 x - 10}{4}$

We can put this into the second equation:

${x}^{2} + {\left(\frac{3 x + 10}{4}\right)}^{2} = {r}^{2}$

${x}^{2} + {\left(3 x + 10\right)}^{2} / {4}^{2} = {r}^{2}$

${x}^{2} + \frac{9 {x}^{2} + 60 x + 100}{16} = {r}^{2}$

$\frac{16 {x}^{2}}{16} + \frac{9 {x}^{2} + 60 x + 100}{16} = {r}^{2}$

$\frac{25 {x}^{2} + 60 x + 100}{16} = {r}^{2}$

$\frac{25}{16} {x}^{2} + \frac{60}{16} x + \frac{100}{16} = {r}^{2}$

$\frac{25}{16} {x}^{2} + \frac{15}{4} x + \frac{25}{4} = {r}^{2}$

Now we have to think a little. The word "tangent" in the question would mean that the system of equations has one solution for $x$ and $y$. Therefore, we want this above quadratic to have one solution.

In order for a quadratic to have exactly one solution (and not two or zero), the discriminant has to be $0$. We can identify the $a$, $b$, and $c$ values of the quadratic, and set the discriminant (${b}^{2} - 4 a c$) equal to $0$.

${\underbrace{\frac{25}{16}}}_{a} {x}^{2} + {\underbrace{\frac{15}{4}}}_{b} x + {\underbrace{\frac{25}{4} - {r}^{2}}}_{c} = 0$

Setting the discriminant equal to $0$:

${\left(\frac{15}{4}\right)}^{2} - 4 \left(\frac{25}{16}\right) \left(\frac{25}{4} - {r}^{2}\right) = 0$

$\frac{225}{16} - \frac{25}{4} \left(\frac{25}{4} - {r}^{2}\right) = 0$

$\frac{225}{16} - \frac{625}{16} + \frac{25}{4} {r}^{2} = 0$

$- \frac{400}{16} + \frac{25}{4} {r}^{2} = 0$

$\frac{25}{4} {r}^{2} = \frac{400}{16}$

$25 {r}^{2} = \frac{1600}{16}$

$25 {r}^{2} = 100$

${r}^{2} = 4$

$r = 2$

We found the radius! We can now write the equation of the circle:

${x}^{2} + {y}^{2} = {2}^{2}$

${x}^{2} + {y}^{2} = 4$

That's the solution. Hope this helped!