# A circle touches two perpendicular lines 2x-3y=15 and 3x+2y=3 at the points A(6,-1), B(1,0) respectively. Find the equation of the circle?

Mar 31, 2018

${x}^{2} + {y}^{2} - 8 x - 4 y + 7 = 0$

#### Explanation:

The tangent to a circle

${x}^{2} + {y}^{2} + 2 a x + 2 b y + c = 0$

at the point $\left({x}_{1} , {y}_{1}\right)$ is given by

$x {x}_{1} + y {y}_{1} + a \left(x + {x}_{1}\right) + b \left(y + {y}_{1}\right) + c = 0$

Thus, the tangent at $A = \left(6 , - 1\right)$ is

$6 x - y + a \left(x + 6\right) + b \left(y - 1\right) + c = 0$

or

$\left(6 + a\right) x + \left(b - 1\right) y + \left(6 a - b + c\right) = 0$

Since this is the line $2 x - 3 y = 15$, we must have

$\frac{6 + a}{2} = \frac{b - 1}{- 3} = \frac{6 a + b - c}{- 15}$

Again, the tangent at $B = \left(1 , 0\right)$ is

$1 x + 0 \times y + a \left(x + 1\right) + b \left(y + 0\right) + c = 0$

or

$\left(1 + a\right) x + b y + \left(a + c\right) = 0$

Since this is the line $3 x + 2 y = 3$, we must have

$\frac{1 + a}{3} = \frac{b}{2} = \frac{a + c}{- 3}$

Hence $b = \frac{2}{3} \left(1 + a\right)$ and substituting this in $\frac{6 + a}{2} = \frac{b - 1}{- 3}$ gives

$\frac{6 + a}{2} = \frac{\frac{2}{3} \left(1 + a\right) - 1}{- 3} = - \frac{2}{9} a + \frac{1}{9} \implies \left(\frac{1}{2} + \frac{2}{9}\right) a = \frac{1}{9} - 3$

Thus

$\frac{13}{18} a = - \frac{26}{9} \implies a = - 4$

So,

$b = \frac{2}{3} \left(a + 1\right) = - 2$

and

$\frac{b}{2} = \frac{a + c}{- 3} \implies - 4 + c = - \frac{3}{2} \times \left(- 2\right) = 3 \implies c = 7$

Thus the circle is

${x}^{2} + {y}^{2} - 8 x - 4 y + 7 = 0$

Mar 31, 2018

${\left(x - 4\right)}^{2} + {\left(y - 2\right)}^{2} = {\left(\sqrt{13}\right)}^{2}$

#### Explanation:

The following is a graph of:

$\textcolor{red}{2 x - 3 y = 15} \text{ [1]}$

$\textcolor{b l u e}{3 x + 2 y = 3} \text{ [2]}$

$\textcolor{g r e e n}{A \left(6 , - 1\right)}$, and $\textcolor{\mathmr{and} a n \ge}{B \left(1 , 0\right)}$

Because the center must be located on lines that are perpendicular to the points of tangency, we can use equations [1] and [2] to write two equations that must intersect at the center.

Set equation [1] equal to an arbitrary constant, ${c}_{1}$:

$2 x - 3 y = {c}_{1} \text{ [1.1]}$

Set equation [2] equal to an arbitrary constant ${c}_{2}$:

$3 x + 2 y = {c}_{2} \text{ [2.1]}$

Substitute point B into equation [1.1] and solve for ${c}_{1}$

$2 \left(1\right) - 3 \left(0\right) = {c}_{1}$

${c}_{1} = 2$

Substitute the above into equation [1.1]:

$2 x - 3 y = 2 \text{ [1.2]}$

Substitute point A into equation [2.1] and solve for ${c}_{2}$:

$3 \left(6\right) + 2 \left(- 1\right) = {c}_{2}$

${c}_{2} = 16$

Substitute the above into equation [2.1]:

$3 x + 2 y = 16 \text{ [2.2]}$

Add equations [1.2] and [2.2] to the graph:

Please understand that the center of the circle must be at the intersection of equations [1.2] and [2.2], therefore, we shall solve them as a system of equations:

$2 x - 3 y = 2 \text{ [1.2]}$
$3 x + 2 y = 16 \text{ [2.2]}$

$4 x - 6 y = 4 \text{ [1.3]}$
$9 x + 6 y = 48 \text{ [2.2]}$

$13 x = 52$

$x = 4$

$3 \left(4\right) + 2 y = 16$

$2 y = 4$

$y = 2$

The center of the circle is at the point $C \left(4 , 2\right)$

The standard Cartesian equation of a circle is:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2} \text{ [3]}$

where $\left(x , y\right)$ is any point on the circle, $\left(h , k\right)$ is the center, and $r$ is the radius.

Substitute point C into equation [3]:

${\left(x - 4\right)}^{2} + {\left(y - 2\right)}^{2} = {r}^{2} \text{ [3.1]}$

To find the radius, we shall substitute point B into equation [3.1]:

${\left(1 - 4\right)}^{2} + {\left(0 - 2\right)}^{2} = {r}^{2}$

$9 + 4 = {r}^{2}$

$13 = {r}^{2}$

$r = \sqrt{13}$

Substitute the above into equation [3.1]:

${\left(x - 4\right)}^{2} + {\left(y - 2\right)}^{2} = {\left(\sqrt{13}\right)}^{2} \text{ [3.2]}$

Add equation [3.2] to the graph:

Please observe that equation [3.2] touches [1] and [2] and points A and B respectively, therefore, equation [3.2] is the correct equation of the circle.