Question

What is the equation of the circle passing through (-4, -4) and tangent to the line 2x – 3y + 9 = 0 at (-3, 1)?

Answer 1

Calculus? Algebra?

Answer 2

Calculus

Answer 3

So we know we have a circle...
\[(x-h)^2+(y-k)^2=r^2 \]

Answer 4

we need to differentiate this

Answer 5

we know to do this because we are given something about a tangent line to the circle
so we are hoping to use that

Answer 6

2x-3y+9=0
-3y=-2x+9
y=2 x/ 3 - 3
slope is 2/3
you check me on that if you want...
But can you differentiate the equation I gave you w.r.t to x
h,k, and r are constants

Answer 7

so what will i do if i found the slope of the tangent line? I don't get it.

Answer 8

y'=2/3
but you need to find y' first

Answer 9

why y'? m=slope=2/3

Answer 10

the slope of tangent line at x=a is the derivative of the function/relation evaluated at x=a

Answer 11

by the way let me fix something
2x-3y+9=0
-3y=-2x-9
y=2x/3+3
slope is still 2/3 for this line
we want to find y' and evaluate y' at the point (-3,1)

Answer 12

can you do that?

Answer 13

i'll try to analyze

Answer 14

\[\frac{d}{dx}(x-h)^2=? \\ \frac{d}{dx}(y-k)^2=? \\ \frac{d}{dx}(r^2)=?\]

Answer 15

Can you compute these derivatives?

Answer 16

I will do the second one for you
\[\frac{d}{dx}(y-k)^2=2(y-k)y'\]

Answer 17

the first one is very similar
and you should know the last one is pretty easy because derivative of a constant is?

Answer 18

ohhh our professor haven't teach us that yet

Answer 19

in specific the lessons is analytic geometry

Answer 20

but our class is calculus 1

Answer 21

ok but we aren't suppose to use calculus right?
just algebra and geometry ?
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\[y=\frac{2}{3}x+3 \text{ is the tangent line of the circle at } (-3,1) \\ \text{ I want to determine the radius } \\ \text{ pretending the center is } (h,k) \\ \text{ We can draw a line from } (h,k) \text{ \to } (-3,1) \\ \text{ We know this line is perpendicular to the tangent line } \\ y=\frac{-3}{2}x+B \\ \text{ we can determine B by using our point } (-3,1) \\ 1=\frac{-3}{2}(-3)+B \\ 1=\frac{9}{2}+B \\ \text{ solve for B to find the y-intercept of this tangent line }\]