Find the equation of the circle passing through A and B

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In summary, the conversation discusses finding the equation of a circle passing through two given points, A(8,1) and B(7,0). The general equation for a circle is provided, and it is explained that three points are needed to fully define the circle. The conversation also mentions finding the equation of the tangent at point B and a parallel tangent. It is concluded that the problem is solvable and three equations are needed to find the center and radius of the circle.
  • #1
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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.
 
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  • #2
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
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
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
thank you.
 

1. What is the equation of a circle?

The equation of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle and r is the radius.

2. How do you find the equation of a circle given two points?

To find the equation of a circle passing through two given points, A(x1, y1) and B(x2, y2), you can use the formula (x - h)^2 + (y - k)^2 = r^2 and plug in the coordinates of the two points into the equation. Then, you can solve for the center (h, k) and radius r using algebraic methods.

3. Can you find the equation of a circle with only one point given?

No, the equation of a circle requires at least two points to be determined. However, if the point given is the center of the circle, then the equation can be written as (x - x1)^2 + (y - y1)^2 = r^2, where (x1, y1) is the center and r is the radius.

4. What if the two points given are collinear?

If the two points given are collinear, then there is no unique circle passing through them. In this case, the equation (x - h)^2 + (y - k)^2 = r^2 represents a circle with infinite radius and center at (h, k) which lies on the line passing through the two points.

5. Are there any special cases when finding the equation of a circle passing through two points?

Yes, there are two special cases: when the two points given are the same point, in which case the equation will be (x - x1)^2 + (y - y1)^2 = 0; and when the two points given are equidistant from the origin, in which case the equation will be x^2 + y^2 = (r/2)^2, where r is the distance between the two points.

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