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

• denian
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.

#### denian

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.

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?

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.

## 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.