# What is the equation of a circle of radius 1, which is also tangent to the line 3x-4y+1=0 at the y-axis value = 1 ?

Jun 19, 2015

There are two possible circles, either side of the line:

${\left(x - \frac{2}{5}\right)}^{2} + {\left(y - \frac{9}{5}\right)}^{2} - 1 = 0$

${\left(x - \frac{8}{5}\right)}^{2} + {\left(y - \frac{1}{5}\right)}^{2} - 1 = 0$

#### Explanation:

When $y = 1$ the equation becomes $3 x - 4 + 1 = 0$, hence $x = 1$.

So the circle is tangential to the line at $\left(1 , 1\right)$.

Starting with $3 x - 4 y + 1 = 0$, add $4 y$ to both sides to get:

$4 y = 3 x + 1$

Divide both sides by $4$ to get:

$y = \frac{3}{4} x + \frac{1}{4}$

The slope of this line is $m = \frac{3}{4}$, so the slope of the radius of the tangential circle passing through the point of contact will be $- \frac{1}{m} = - \frac{4}{3}$.

So this perpendicular line will have equation:

$y = - \frac{4}{3} x + \frac{7}{3}$ in order that it passes through $\left(1 , 1\right)$

The centre of the tangential circle lies at distance $1$ from $\left(1 , 1\right)$ along this line.

That give two possibilities:

$\left(1 - \frac{3}{5} , 1 + \frac{4}{5}\right) = \left(\frac{2}{5} , \frac{9}{5}\right)$

and

$\left(1 + \frac{3}{5} , 1 - \frac{4}{5}\right) = \left(\frac{8}{5} , \frac{1}{5}\right)$

...using the right angled property of the 3, 4, 5 triangle.

The corresponding circles have equations:

${\left(x - \frac{2}{5}\right)}^{2} + {\left(y - \frac{9}{5}\right)}^{2} - 1 = 0$

${\left(x - \frac{8}{5}\right)}^{2} + {\left(y - \frac{1}{5}\right)}^{2} - 1 = 0$

graph{(3x-4y+1)((x-2/5)^2 + (y-9/5)^2 - 1)((x-8/5)^2 + (y - 1/5)^2 - 1) = 0 [-3.52, 6.48, -1.28, 3.72]}