# 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:

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

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

$\textcolor{g r e e n}{A \left(6 , - 1\right)}$, 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  and  to write two equations that must intersect at the center.

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

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

Set equation  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:

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

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 :

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

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  and  and points A and B respectively, therefore, equation [3.2] is the correct equation of the circle.