# Find Equation of Circle: A(2,2) & B(5,3), y=x+1

In summary, the equation of the circle with center (2,2) and radius of 3 is (x-2)^2 + (y-2)^2 = 3^2. To find the center and radius of a circle given two points on its circumference, the center is the midpoint of the line segment connecting the two points and the radius is the distance between the center and either of the two points. The equation of the circle can be written in standard form. To determine if the circle intersects with a given line, we can substitute the line's equation into the circle's equation and check for real solutions. The equation of the tangent line to the circle at the point (5,3) is y = -2x
how would you find the equation of a circle : (x-p)^2 + (y-q)^2=r^2 with this given info:

the circle passes through A(2,2) and B(5,3) and its centre is on the line defined by y=X+1

Plug the two points in the equation and then you get an equation with p and q which should look like 3p+2q=13.

since the center is on the line y=x+1, and the center of the circle is supposed to be (p,q). So q=p+1.

then solve for the equation.

To find the equation of a circle, we need to know the coordinates of its center and its radius. In this case, we are given that the center of the circle lies on the line defined by y=x+1. This means that the x-coordinate and y-coordinate of the center must satisfy the equation y=x+1.

We also know that the circle passes through points A(2,2) and B(5,3). This means that the distance from the center of the circle to these two points must be equal to the radius of the circle.

To find the center of the circle, we can use the midpoint formula, which states that the coordinates of the midpoint of a line segment with endpoints (x1, y1) and (x2, y2) are given by ( (x1+x2)/2, (y1+y2)/2). In this case, the midpoint of the line segment AB is ((2+5)/2, (2+3)/2) = (3.5, 2.5).

Now, to find the radius of the circle, we can use the distance formula, which states that the distance between two points (x1, y1) and (x2, y2) is given by √((x2-x1)^2 + (y2-y1)^2). In this case, the distance between the center of the circle and point A is √((2-3.5)^2 + (2-2.5)^2) = √(2.25 + 0.25) = √2.5. Similarly, the distance between the center and point B is √((5-3.5)^2 + (3-2.5)^2) = √(2.25 + 0.25) = √2.5. Therefore, the radius of the circle is √2.5.

Now, we can plug in the values of the center and radius into the general equation of a circle, (x-p)^2 + (y-q)^2=r^2, to get the equation of the circle passing through A(2,2) and B(5,3) with its center on the line y=x+1. Substituting p=3.5, q=2.5, and r=√2.5, we get the equation (x-3.5

## 1. What is the equation of the circle with center (2,2) and radius of 3?

The equation of a circle with center (h,k) and radius r is (x-h)^2 + (y-k)^2 = r^2. Plugging in the values from the given coordinates, the equation of the circle is (x-2)^2 + (y-2)^2 = 3^2.

## 2. How do I find the center and radius of a circle given two points on its circumference?

The center of a circle is the midpoint of the line segment connecting the two points on its circumference. The radius is the distance between the center and either of the two points. In this case, the center is the midpoint of the line segment connecting (2,2) and (5,3), which is ((2+5)/2, (2+3)/2) = (3.5, 2.5). The radius can be found by using the distance formula between the center and one of the points, which is sqrt((3.5-2)^2 + (2.5-2)^2) = sqrt(2.25 + 0.25) = sqrt(2.5).

## 3. Can the equation of the circle be written in standard form?

Yes, the equation of the circle can be written in standard form, which is (x-h)^2 + (y-k)^2 = r^2. In this case, the standard form of the equation is (x-2)^2 + (y-2)^2 = 3^2.

## 4. How do I know if the circle intersects with the line y=x+1?

To determine if the circle intersects with the line y=x+1, we can substitute the equation of the line into the equation of the circle and check if there are any real solutions. In this case, when we substitute y=x+1 into the equation of the circle, we get (x-2)^2 + (x+1-2)^2 = 3^2, which simplifies to 2x^2 - 6x + 2 = 0. This quadratic equation has two real solutions, indicating that the circle and the line intersect at two points.

## 5. What is the equation of the tangent line to the circle at the point (5,3)?

The tangent line to a circle at a given point is perpendicular to the radius of the circle at that point. Therefore, the slope of the tangent line is the negative reciprocal of the slope of the radius, which is -(5-3)/(3-2) = -2/1 = -2. Using the point-slope form of a line, the equation of the tangent line is y-3 = -2(x-5), which can be simplified to y = -2x + 13.

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