# How do you write the equation of the circle passing through (−2,1) and tangent to the line 3x − 2y = 6 at the point (4, 3)?

Dec 1, 2016

${\left(x + \frac{2}{7}\right)}^{2} + {\left(y - \frac{41}{7}\right)}^{2} = \frac{1300}{49}$

#### Explanation:

The circle equation is

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

Now it pass by the points ${p}_{1} = \left(- 2 , 1\right)$ and ${p}_{2} = \left(4 , 3\right)$

and also is tangent to the straight

$L \to 3 x - 2 y - 6 = 0$

So we have that for $C$

$2 \left(x - {x}_{0}\right) \mathrm{dx} + 2 \left(y - {y}_{0}\right) \mathrm{dy} = 0$ or
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x - {x}_{0}}{y - {y}_{0}}$

and also for $L$

$3 \mathrm{dx} - 2 \mathrm{dy} = 0$ or
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{2}$

We also know that at ${p}_{2}$

$- \frac{{x}_{2} - {x}_{0}}{{y}_{2} - {y}_{0}} = \frac{3}{2}$ so

{((x_1-x_0)^2+(y_1-y_0)^2= r^2), ((x_2-x_0)^2+(y_2-y_0)^2=r^2), (-2(x_2-x_0)=3(y_2-y_0)):}

Here the variables to find are ${x}_{0} , {y}_{0} , r$

Subtracting the first and the second we have a new reduced system

$\left\{\begin{matrix}{x}_{1}^{2} - {x}_{2}^{2} + {y}_{1}^{2} - {y}_{2}^{2} + 2 {x}_{0} \left({x}_{2} - {x}_{1}\right) + 2 {y}_{0} \left({y}_{2} - {y}_{1}\right) = 0 \\ - 2 \left({x}_{2} - {x}_{0}\right) = 3 \left({y}_{2} - {y}_{0}\right)\end{matrix}\right.$ with solutions

${x}_{0} = \frac{3 {x}_{1}^{2} - 3 {x}_{2}^{2} + 3 {\left({y}_{1} - {y}_{2}\right)}^{2} + 4 {x}_{2} \left({y}_{2} - {y}_{1}\right)}{6 \left({x}_{1} - {x}_{2}\right) - 4 {y}_{1} + 4 {y}_{2}} = - \frac{2}{7}$

${y}_{0} = \frac{{\left({x}_{1} - {x}_{2}\right)}^{2} + {y}_{1}^{2} + 3 \left(- {x}_{1} + {x}_{2}\right) {y}_{2} - {y}_{2}^{2}}{3 \left({x}_{2} - {x}_{1}\right) + 2 \left({y}_{1} - {y}_{2}\right)} = \frac{41}{7}$

finally

${r}^{2} = {\left({x}_{1} - {x}_{0}\right)}^{2} + {\left({y}_{1} - {y}_{0}\right)}^{2}$ giving

$r = \frac{10 \sqrt{13}}{7}$ Dec 1, 2016

Another approach to the problem is to consider that the circle's center ${p}_{0}$ is located at the intersection of two lines: ${p}_{0} = {L}_{1} \cap {L}_{2}$ where

${L}_{1} \to {\left\mid p - {p}_{1} \right\mid}^{2} = {\left\mid p - {p}_{2} \right\mid}^{2}$
${L}_{2} \to p = {p}_{2} + \lambda \vec{v}$

where $p = \left(x , y\right) , {p}_{1} = \left(- 2 , 1\right)$ and ${p}_{2} = \left(4 , 3\right)$

From $L \to 3 x - 2 y - 6 = 0$ we get $\vec{v} = \left(2 , 3\right)$

the parameter $\lambda \in \mathbb{R}$

Expanding ${L}_{1}$ we get

${L}_{1} \to < p , {p}_{2} - {p}_{1} > = \frac{1}{2} \left({\left\mid {p}_{2} \right\mid}^{2} - {\left\mid {p}_{1} \right\mid}^{2}\right)$

substituting $p$ from ${L}_{2}$ into ${L}_{1}$ we get

$< {p}_{2} + {\lambda}_{0} \vec{v} , {p}_{2} - {p}_{1} > = {\left\mid {p}_{2} \right\mid}^{2} - < {p}_{2} , {p}_{1} > + \lambda < \vec{v} , {p}_{2} - {p}_{1} >$ so

${\lambda}_{0} = - \frac{{\left\mid {p}_{2} \right\mid}^{2} + {\left\mid {p}_{1} \right\mid}^{2} + 2 < {p}_{2} , {p}_{1} >}{< \vec{v} , {p}_{2} - {p}_{1} >}$

then

${p}_{0} = {p}_{2} + {\lambda}_{0} \vec{v}$ and also

$r = \left\mid {p}_{0} - {p}_{2} \right\mid$