# Write the equation of a circle with centre (3,-1) which is tangent to the line y=x+2?

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Nimo N. Share
Feb 23, 2018

See below.

#### Explanation:

Question:
Write the equation of a circle
with centre  color(blue)( C = (3, -1) , which is
tangent to the line  color(blue)( L: " " y = x + 2  ?

To start, we can use the general form of circle equation:
 color(blue)( (x – h)^2 + (y – k)^2 = r^2 .
Center located at:  color(blue)( C = (h, k) ;
Radius:  color(blue)( r .

Doing a little planning.
We need to find the distance between the line, L, and the center, C.

That distance is to be measured along a line perpendicular to L and passing through C, since the radius of a circle is perpendicular to a tangent line at the point of contact.

• The definition of the distance between a point and a line is the measure of a segment drawn from the point perpendicular to the line.

The perpendicular line, ${L}_{\text{perp }}$, has slope that is the negative reciprocal of the slope of the line.
In our case, this line must pass thorough $C = \left(3 , - 1\right)$.
L has slope 1, so the slope of ${L}_{\text{perp }}$ is -1.

Point-slope form:
${L}_{\text{perp": " }} \left(y - h\right) = - 1 \cdot \left(x - k\right)$
${L}_{\text{perp": " }} \left(y - \left(- 1\right)\right) = - 1 \cdot \left(x - \left(3\right)\right)$
${L}_{\text{perp": " }} y + 1 = - 1 x + 3$
Subtract 1 from each side.
 color(brown)( L_"perp": " " y = -x + 2

P will be located at the intersection of ${L}_{\text{perp}}$ and L.
$- x + 2 = x + 2$, so
$x = 0$.
Find the y-coordinate for P.
$y = \left(0\right) + 2 = 2$

$P = \left(0 , 2\right)$

Calculate the distance between P and C and use it to complete the formula for the circle.
{This is the radius of the circle.}
$r : \text{ } \left\mid \overline{C P} \right\mid = \sqrt{{\left(3 - 0\right)}^{2} + {\left(- 1 - 2\right)}^{2}}$
 color(brown)( r: " "abs(bar(CP)) = sqrt( 9 + 9 ) = sqrt( 18 )
We don't need to calculate the number, since we are going to use ${r}^{2}$ in the circle formula.

We are ready for the final result.
 color(blue)( (x – h)^2 + (y – k)^2 = r^2 .
 (x – (3))^2 + (y – (-1))^2 = 18 .
 color(brown)( (x – 3)^2 + (y + 1)^2 = 18 .

Graph.

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Feb 23, 2018

${\left(x - 3\right)}^{2} + {\left(y + 1\right)}^{2} = 18$

#### Explanation:

Given -
Centre of the circle $\left(3 , - 1\right)$
Equation of the tangent line -$y = x + 2$ ----------- (1)

We have to find the radius of the circle.

Find the equation of the line passing through point of tangency and the centre. It cuts the tangent at 90 degrees.

The slope of the tangent is ${m}_{1} = 1$
The slope of the required line is ${m}_{2} = \frac{- 1}{{m}_{1}} = \frac{- 1}{1} = - 1$

Fond the equation of the required line

$m x + c = y$
$- 1 \left(3\right) + c = - 1$
$- 3 + c = - 1$
$c = - 1 + 3 = 2$
$c = 2$

$y = - x + 2$ ---------------(2)

Find the point where both the lines cut each other.

$y = x + 2$ --------------- (1)
$y = - x + 2$ ---------------(2) add both

$2 y = 4$
$y = \frac{4}{2} = 2$

Substitute $y = 2$ in equation (1)

$2 = x + 2$
$x = 2 - 2 = 0$

The two lines cut at $\left(0 , 2\right)$

Distance between $\left(0 , 2\right)$ and $\left(3 , - 1\right)$ is the radius

${r}^{2} = {\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}$
${r}^{2} = {\left(0 - 3\right)}^{2} + {\left(2 - \left(- 1\right)\right)}^{2}$
${r}^{2} = {\left(- 3\right)}^{2} + {\left(2 + 1\right)}^{2} = 9 + 9 = 18$

We have Centre and radius squared

The equation is -

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

${\left(x - 3\right)}^{2} + {\left(y + 1\right)}^{2} = 18$

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Alan N. Share
Feb 23, 2018

${\left(x - 3\right)}^{2} + {\left(y + 1\right)}^{2} = 18$

#### Explanation:

The equation of a circle centred at the point $\left(3 , - 1\right)$ has the equation:

${\left(x - 3\right)}^{2} + {\left(y + 1\right)}^{2} = {r}^{2}$
Where $r$ is the radius of the circle.

Hence, this question asks us to find ${r}^{2}$.

We know that the line $y = x + 2$ is a tangent to this circle.

Note that this line has a slope of $1$

We also know that the slope of any circle centred at $\left({x}_{1} , {y}_{1}\right)$ at the point $\left(x , y\right)$ is given by: $- \frac{y - {y}_{1}}{x - {x}_{1}}$; which in our case is:$- \frac{y + 1}{x - 3}$

Since the tangent touches the circle at a point where the circle has a slope of 1, we can replace $y = x + 2$ in the slope function and set this equal to 1,

$\therefore \frac{- \left(x + 2 + 1\right)}{x - 3} = 1$

$- x - 3 = x - 3$

$- 2 x = 0 \to x = 0$ at the point of intersection

Now, since $y = x + 2 \to y = 2$ at the point of intersection.

Finally, we can arrive at ${r}^{2}$ by applying the point $\left(0 , 2\right)$ to the equation of our circle.

${\left(x - 3\right)}^{2} + {\left(y + 1\right)}^{2} = {r}^{2}$ at $\left(0 , 2\right) \to$

${\left(- 3\right)}^{2} + {\left(+ 3\right)}^{2} = {r}^{2}$

${r}^{2} = 9 + 9 = 18$

Hence, the equation of our circle is: ${\left(x - 3\right)}^{2} + {\left(y + 1\right)}^{2} = 18$

This is depicted graphically below.

graph{((x-3)^2+(y+1)^2-18)(x+2-y)=0 [-18.34, 17.68, -6.8, 11.23]}

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