Circles

1) Equation of circle given center and radius:

The equation of a circle with center (a,b) and radius r is given by (x-a)² + (y-b)² = r².
This equation can be derived from the distance formula, by setting the distance between (x,y) and (a,b) equal to r.
The center of the circle is the point (a,b), and the radius is r.
The equation can be written in general form as x² + y² + Dx + Ey + F = 0, where D = -2a, E = -2b, and F = a² + b² - r².
The standard form of the equation is (x-h)² + (y-k)² = r², where (h,k) is the center of the circle.

The general equation of a circle is given by Ax² + Ay² + Bx + Cy + D = 0, where A and B are not both zero.
This equation can be obtained by substituting x² and y² with their respective coefficients in the standard form of the equation.
The coefficients A, B, and C determine the position of the circle in the xy-plane, while D determines its size.
To convert the general equation to standard form, complete the square for both x and y, and then combine the resulting expressions.
The standard form of the equation can be used to find the center and radius of the circle.

To find the center and radius of a circle given its equation in standard form, rewrite the equation in the form (x-h)² + (y-k)² = r².
Complete the square for both x and y, and then rearrange the terms to isolate the center (h,k) and the radius r.
The center of the circle is (h,k), and the radius is r.
To find the center and radius of a circle given its equation in general form, convert the equation to standard form first, and then use the method described above.
If the equation of the circle is given in another form, such as center and a point on the circumference or 3 non-collinear points, use the appropriate method to find the center and radius.

To find the center and radius of a circle given its center (h,k) and a point on the circumference (x1,y1), use the distance formula.
The distance between the center and the point on the circumference is equal to the radius of the circle.
The center of the circle is the midpoint of the line segment joining (h,k) and (x1,y1), and the radius is the distance between the center and (x1,y1).
Alternatively, use the equation of the circle to solve for the radius, and then substitute the coordinates of the point on the circumference to solve for the center.
There are two possible circles that can pass through a given center and point on the circumference, one with a positive radius and one with a negative radius.

To find the center and radius of a circle given three non-collinear points (x1,y1), (x2,y2), and (x3,y3), use the circumcenter formula.
The circumcenter is the point of intersection of the perpendicular bisectors of the sides of the triangle formed by the three points.
Find the slope and midpoint of each side of the triangle, and then find the equations of the perpendicular bisectors.
The point of intersection of the perpendicular bisectors is the center of the circle.
The radius of the circle is the distance between the center and any of the three points.

Center: $$(a,b)$$
Points: $$(x_1,y_1), (x_2,y_2), (x_3,y_3)$$
Equation: $$\begin{aligned} a &= \frac{1}{2}\begin{vmatrix} x_1^2 + y_1^2 & y_1 & 1 \\ x_2^2 + y_2^2 & y_2 & 1 \\ x_3^2 + y_3^2 & y_3 & 1 \end{vmatrix} \\ b &= -\frac{1}{2}\begin{vmatrix} x_1^2 + y_1^2 & x_1 & 1 \\ x_2^2 + y_2^2 & x_2 & 1 \\ x_3^2 + y_3^2 & x_3 & 1 \end{vmatrix} \\ r &= \sqrt{(x_1 - a)^2 + (y_1 - b)^2} \end{aligned}$$

To find the center and radius of a circle given its center (h,k) and a tangent line at a point (x1,y1) on the circumference, use the perpendicular bisector of the tangent line.
The perpendicular bisector of the tangent line passes through the center of the circle.
Find the slope of the tangent line and the midpoint of the line segment joining (h,k) and (x1,y1), and then find the equation of the perpendicular bisector.
The point of intersection of the perpendicular bisector and the line passing through (h,k) and (x1,y1) is the center of the circle.
The radius of the circle is the distance between the center and (x1,y1).

The tangent line to a circle at a point (x1,y1) on the circumference is perpendicular to the radius passing through the point.
The slope of the tangent line is equal to the negative reciprocal of the slope of the radius passing through the point.
The equation of the tangent line is y - y1 = m(x - x1), where m is the slope of the tangent line.
The normal line to the circle at the same point is perpendicular to the tangent line and passes through the point (x1,y1).
The slope of the normal line is equal to the negative of the slope of the tangent line, and the equation of the normal line is y - y1 = (-1/m)(x - x1), where m is the slope of the tangent line.

Center: $$(a,b)$$
Point on circle: $$(x_1,y_1)$$
Equation of tangent: $$\frac{x(x_1 - a) + y(y_1 - b)}{r^2} = \frac{x_1 - a}{r} = \frac{y_1 - b}{r}$$
Equation of normal: $$\frac{x(a - x_1) + y(b - y_1)}{r^2} = \frac{a - x_1}{r} = \frac{b - y_1}{r}$$