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

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

asked Aug 18, 2021 66.8k views

1 Answer

Answer:

So the points of tangency are:

(-2 + (3√(2))/(2),3+(3√(2))/(2))

and

(-2 - (3√(2))/(2),3-(3√(2))/(2))

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
.

Subtract
2(x+2) on both sides:

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

Divide both sides by
2(y-3) :

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

Simplify:

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

We want to find the points such that:

y'=-1

(-(x+2))/(y-3)=-1

Multiply both sides by
-(y-3) :

(x+2)=y-3

Add
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
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=(9)/(2)

Take square root of both sides:

$x+2=\pm \sqrt{(9)/(2)}$

$x+2=\pm (√(9))/(√(2))$

$x+2=\pm (3)/(√(2))$

Rationalize denominator by multiplying top and bottom by
√(2) for right hand side:

$x+2=\pm (3√(2))/(2)$

Subtract 2 on both sides:

$x=-2 \pm (3√(2))/(2)$

Now let's find the corresponding
coordinates.

y=x+5

$y=(-2 \pm (3√(2))/(2))+5$

$y=(-2+5) \pm (3√(2))/(2)$

$y=3 \pm (3√(2))/(2)$

So the points of tangency are:

(-2 + (3√(2))/(2),3+(3√(2))/(2))

and

(-2 - (3√(2))/(2),3-(3√(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
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.

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

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

Peter Monks answered Aug 22, 2021

7.3k points

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