Question

Find the equation of the circle that passes through (2,3) is tangent to the line 3x-4 = -1

Answer 1

Here's what i've done so far

\[(x-h)^{2}+(y-k)^{2} = r \]

\[\frac{ d }{ dx }(x-h)^{2}+\frac{ d }{ dx }(y-k)^{2} = 0 \]

\[\frac{ dy }{ dx }2(y-k) = -2(x+h)\]

\[\frac{ dy }{ dx } = \frac{ -(x+h) }{ (y-k) }\]

\[3x-4y = -1 \]

\[3x= -1+4y \]

\[3x+1 = 4y \]

\[\frac{ 3 }{ 4 }x+\frac{ 1 }{ 4 } = y\]

since the slope of the tangent line is 3/4 i set this equal to the derivative of the circle

\[\frac{ -(x-h) }{ (y-k) } = \frac{ 3 }{ 4 }\]

Answer 2

Now I thought that you just plug in the points x = 2 y = 3 and solve for h and k

\[\frac{ -(2-h) }{ (3-k) } = \frac{ 3 }{ 4 } ~h = 5, k = -1 \]

\[(x-5)^{2}+(y+1)^{2} = 25 \]

Answer 3

Your circle goes through (2,3) just fine, but `3x-4y=-1` is not tangent to this circle.

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Answer 4

argh.. I missed something big... the question is actually find the standard equations of the circles that pass through (2,3) and are tangent to both lines (3x-4y) = -1 and 4x+3y = 7..

Answer 5

that's the question word for word. not sure here.

Answer 6

so the circle would look something like this

Answer 7

I would set up points P and Q in the form of (x,y)

solve each equation for y to get it in terms of x

Answer 8

so I would need two separate equations

Answer 9

So, would one equation be this?

\[\frac{ -(x-h) }{ (y-k) } = \frac{ 3 }{ 4 }\]

\[-\frac{ 4 }{ 3 }*(x-h)+k = y_{1}\]


\[\frac{ -(x-h) }{ (y-k) } = \frac{ -4}{3} = -\frac{ 3 }{ 4 }*(x-h)+k = y_{2}\]

Answer 10

I'm wondering if i'm on the right track with this

Answer 11

http://assets.openstudy.com/updates/attachments/572d45d3e4b071c0ef3c0e92-jim_thompson5910-1462586209770-circle2.png

focus on the purple equation 4x+3y = 7

solve for y to get

\[\Large y = \frac{7-4x}{3}\]

so if P is some point on this purple equation, and P is allowed to slide anywhere along this line, then point P is

\[\Large P = \left(x, \frac{7-4x}{3}\right)\]

Agreed?

Answer 12

yes I'm following

Answer 13

similarly, we solve the red equation 3x - 4y = -1 for y to get

\[\Large y = \frac{1+3x}{4}\]

making

\[\Large Q = \left(x, \frac{1+3x}{4}\right)\]

where Q is any point on the red line

Agreed?

Answer 14

yep

Answer 15

so we have some point R that is the center of this circle

Answer 16

following

Answer 17

we don't know what point R is, but the goal is to make

slope of segment RP perpendicular to the purple line
slope of segment RQ perpendicular to the red line

Answer 18

we don't know what R is, but we know its the center of the circle, so

R = (h,k)

Answer 19

`slope of segment RP perpendicular to the purple line`

so

slope of RP = 3/4

since slope of purple line = -4/3 and you flip the fraction and the sign to get the perpendicular slope

Answer 20

`slope of segment RQ perpendicular to the red line`

so

slope of RQ = -4/3

since slope of purple line = 3/4 and you flip the fraction and the sign to get the perpendicular slope

Answer 21

interesting, so you take the slopes of the two lines and take the reciprocal of both.

Answer 22

are you able to find the slope of RP in general (in terms of x)?

hint: slope formula

Answer 23

sorry I meant in terms of x, h, and k

Answer 24

RP = 3/4
RQ = 4/3

Answer 25

RQ has negative slope

Answer 26

So this is what I found for the slope taking the derivative of the general formula for the circle.

\[-\frac{ (x-h) }{ (y-k) }\]

Answer 27

\[RQ = \frac{ -(x-h) }{ (y-k)} = \frac{ -4 }{ 3 }\]

\[RP = \frac{ -(x-h) }{ (y-k)} = \frac{ 3}{ 4 }\]

Answer 28

My next idea is to I guess plug in the points P and Q into the two equations and set them equal to each-other

Answer 29

\[\Large R = \left(h,k\right) = (x_1,y_1)\]
\[\Large P = \left(x, \frac{7-4x}{3}\right) = (x_2,y_2)\]
Use the slope formula to find the slope of RP in general (in terms of x,h,k)
\[\Large m = \frac{y_2 - y_1}{x_2 - x_1}\]
\[\Large m = \frac{\frac{7-4x}{3} - k}{x - h}\]

Now plug in the slope of RP (earlier we found it to be 3/4) then isolate the 'x-h' term
\[\Large m = \frac{\frac{7-4x}{3} - k}{x - h}\]
\[\Large \frac{3}{4} = \frac{\frac{7-4x}{3} - k}{x - h}\]
\[\Large 3*(x - h) = 4*\left(\frac{7-4x}{3} - k\right)\]
\[\Large x-h = \frac{4*\left(\frac{7-4x}{3} - k\right)}{3}\]

Answer 30

okay, I'm starting to see this a bit better so you would do that for the other point and then with those two equations you can find the value of h, k

Answer 31

hmm I'm probably making it more complicated than it has to be