# How do you write the equation for a circle with center at (0, 0) and touching the line 3x+4y=10?

Jun 13, 2016

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

#### Explanation:

To find the equation of a circle we should have the center and radius.
Equation of circle is:

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

Where (a,b) : are the coordinates of the center and

Given the center (0,0)

We should find the radius .
Radius is the perpendicular distance between (0,0) and the line 3x+4y=10

Applying the property of the distance $d$ between line $A x + B y + C$ and point $\left(m , n\right)$ that says:

$$     d = |A*m+B*n+C| / sqrt(A^2+B^2)


The radius which is the distance from straight line $3 x + 4 y - 10 = 0$ to the center $\left(0 , 0\right)$ we have:

A=3. B=4 and C=-10
So,

$r =$
$| 3 \cdot 0 + 4 \cdot 0 - 10 \frac{|}{\sqrt{{3}^{2} + {4}^{2}}}$

= $| 0 + 0 - 10 \frac{|}{\sqrt{9 + 16}}$

= $\frac{10}{\sqrt{25}}$
=$\frac{10}{5}$
=$2$

So the equation of the circle of center(0,0) and radius 2 is :

${\left(x - 0\right)}^{2} + {\left(y - 0\right)}^{2} = {2}^{2}$

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