# Find the equation of the circle passing through A and B

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.

jamesrc
Gold Member
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.

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?

jamesrc
Gold Member
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

thank you.