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Find the equation of the circle passing through A and B

  • Thread starter denian
  • Start date
  • #1
160
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point A had the coordinates (8,1) and B(7,0)
find the equation of the circle passing through A and B.

hope to get rough idea on how to solve this question. tq.
 

Answers and Replies

  • #2
jamesrc
Science Advisor
Gold Member
476
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general equation for a circle:

(x-xc)^2 + (y-yc)^2 = R^2

where xc is the x-coordinate of the center of the circle, yc is the y-coordinate of the center of the circle, and R is the radius of the circle.

To fully define a circle, you need to specify 3 points on its circumference (unless you, for example, know the location of the center). This is borne out in the equations: If you plug in the known info to this equation:

point A:

(8-xc)^2 + (1-yc)^2 = R^2

point B:

(7-xc)^2 + (0-yc^2) = R^2


You have 2 equations and 3 unknowns, so you don't have a uniquely defined circle. If you know xc, yc, or R, you can proceed and write out the equation for the circle.
 
  • #3
160
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here's the actual question.

point A had the coordinates (8,1)
and B (7,0). find the equation of the circle passing through A and B, and its tangent at the point B had the equation 3x-4y-21=0.
find the equation of the tangent parallel with the tangent at B.

can it be solved?
 
  • #4
jamesrc
Science Advisor
Gold Member
476
1
Yes, it is solvable.

Get your third equation from the slope of the line:

first write the equation of the line in standard form:
3x-4y-21=0 --> y = .75x-(21/4)

Now find an expression for the slope from the equation for the circle:

(x-xc)^2 + (y-yc)^2 = R^2

use implicit differentiation to get:

2(x-xc)dx + 2(y-yc)dy = 0 --> (dy/dx) = -(x-xc)/(y-yc)

Evaluate this at point B so you can equate with the slope above:

(dy/dx)[at B] = -(7-xc)/(0-yc) = 0.75

so you have your 3 equations:

xc = 7 - .75yc
(8-xc)^2 + (1-yc)^2 = R^2
(7-xc)^2 + (0-yc^2) = R^2

and you can use those to find xc, yc, and R
 
  • #5
160
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thank you.
 

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