Question

(x + 2)^2 + (y - 3)^2 = 9. There are two lines tangent to this circle having a slope of -1

Answer 1

Answer:

So the points of tangency are:

$(-2 + \frac{3\sqrt{2}}{2},3+\frac{3\sqrt{2}}{2})$

and

$(-2 - \frac{3\sqrt{2}}{2},3-\frac{3\sqrt{2}}{2})$

Step-by-step explanation:

$(x+2)^2+(y-3)^2=9$

Since the instructions are to find the points of tangency where the slopes are -1, we will need ti find the derivative.

$2(x+2)^1(x+2)'+2(y-3)^1(y-3)'=0$

I used chain rule on the left and constant rule on the right.

Let's finish differentiating the insides:

$2(x+2)(1+0)+2(y-3)(y'-0)=0$

Simplify:

$2(x+2)(1)+2(y-3)y'=0$

$2(x+2)+2(y-3)y'=0$

Now we want to isolate $y'$.

Subtract $2(x+2)$ on both sides:

$2(y-3)y'=-2(x+2)$

Divide both sides by $2(y-3)$:

$y'=\frac{-2(x+2)}{2(y-3)}$

Simplify:

$y'=\frac{-(x+2)}{y-3}$

We want to find the points such that:

$y'=-1$

$\frac{-(x+2)}{y-3}=-1$

Multiply both sides by $-(y-3)$:

$(x+2)=y-3$

Add $3$ on both sides:

$(x+2)+3=y$

Distribute:

$x+2+3=y$

Simplify:

$x+5=y$

So we want to find the points such that the following two equations are satisfied:

$(x+2)^2+(y-3)^2=9$

$y=x+5$

Let's replace $y$ with $(x+5)$:

$(x+2)^2+((x+5)-3)^2=9$

$(x+2)^2+(x+5-3)^2=9$

$(x+2)^2+(x+2)^2=9$

Combine like terms:

$2(x+2)^2=9$

Divide both sides by 2:

$(x+2)^2=\frac{9}{2}$

Take square root of both sides:

$x+2=\pm \sqrt{\frac{9}{2}}$

$x+2=\pm \frac{\sqrt{9}}{\sqrt{2}}$

$x+2=\pm \frac{3}{\sqrt{2}}$

Rationalize denominator by multiplying top and bottom by $\sqrt{2}$ for right hand side:

$x+2=\pm \frac{3\sqrt{2}}{2}$

Subtract 2 on both sides:

$x=-2 \pm \frac{3\sqrt{2}}{2}$

Now let's find the corresponding $y-$ coordinates.

$y=x+5$

$y=(-2 \pm \frac{3\sqrt{2}}{2})+5$

$y=(-2+5) \pm \frac{3\sqrt{2}}{2}$

$y=3 \pm \frac{3\sqrt{2}}{2}$

So the points of tangency are:

$(-2 + \frac{3\sqrt{2}}{2},3+\frac{3\sqrt{2}}{2})$

and

$(-2 - \frac{3\sqrt{2}}{2},3-\frac{3\sqrt{2}}{2})$

---------------------Adding geometric/algebra approach here------

$(x+2)^2+(y-3)^2=9$ has center (-2,3) and radius 3.

I put a drawing to show sort of what we are looking for.

We can find the point of tangency by finding where the green line is perpendicular to while going through center of circle which is (-2,3).

Perpendicular lines have opposite reciprocal slopes. The opposite reciprocal of -1 is 1/1 =1 (so just 1 is the slope of the line perpendicular to the green line. The purple line that I drew will represent our perpendicular line.

So we are looking for a line with slope 1 and goes through point (-2,3).

Slope-intercept form is $y=mx+b$.

We know $m=1$:

$y=1x+b$

$y=x+b$

We can find $b$ using (-2,3):

$3=-2+b$

Add 2 on both sides:

$3+2=b$

So the line is $y=x+5$ which is the line we found during our calculus approach.