Conics, a family of similarly shaped curves, Circle, General equation of a circle with the center S(p, q) - translated circle, Circle through three points, Circle and Line, Equation of a tangent at a point of a circle with the center at the origin

Analytic geometry - Conic sections

Conics, a family of similarly shaped curves

:: Circle

By intersecting either of the two right circular conical surfaces (nappes) with the plane perpendicular to the axis of the cone the resulting intersection is a circle.

General equation of a circle with the center S(p, q) - translated circle

A circle with the center at the point S(p, q) and radius r is a set of all points P(x, y) of a plane to whom

the distance from the center, SP = r or

is the general equation of a circle with the center S(p, q).

Equation of the circle with the center at the origin O(0, 0),

Example: A circle passes through points A(2, 4) and B(-2, 6) and its center lies on a line x + 3y - 8 = 0. Find equation of the circle.

Solution: The intersection of the chord AB bisector and the given line is the center S of the circle, since the bisector is normal through the midpoint M, then

As the bisector is perpendicular to the line AB

Equation of the bisector

M(0, 5) and m_n = 2 => y - y₁ = m( x - x₁),

gives y - 5 = 2(x - 0) or -2x + y - 5 = 0.

Therefore, the equation of the circle, (x - p)² + (y - q)² = r² => (x + 1)² + (y - 3)² = 10.

Circle through three points

A circle is uniquely determined by three points not lying on the same line. If given are points, A, B and C then the intersection of any pair of the perpendicular bisectors of the sides of the triangle ABC is the center of the circle.

Since all three points lie on the circle, their coordinates must satisfy equation of the circle

(x - p)² + (y - q)² = r².

Thus, we obtain the system of three equations in three unknowns p, q and r.

Subtracting second equation from first and then third from first we obtain two equations in two unknowns p and q.

Solutions of that system plug into any of three equations to get r.

Example: Find equation of a circle passing through three points, A(-2, -6), B(5, -7) and C(6, 0).

Solution: The coordinates of the points, A, B and C plug into equation (x - p)² + (y - q)² = r²,

Thus, the equation of the circle through points A, B and C, (x - 2)² + (y + 3)² = 25.

:: Circle and Line

Line circle intersection

A line and a circle in a plane can have one of the three positions in relation to each other, depending on the distance d of the center S(p, q) of the circle

(x - p)² + (y - q)² = r²from the line Ax + By + C = 0,

where the formula for the distance:

If the distance of the center of a circle from a line is such that:

d < r, then the line intersects the circle in two points,

d = r, the line touches the circle at only one point,

d > r, the line does not intersect the circle, and they have no common points.

Example: At which points the line x + 5y + 16 = 0 intersects the circle x² + y² - 4x + 2y - 8 = 0.

Solution: To find coordinates of points at which the line intersects the circle solve the system of equations

So, the line intersects the circle at points, A(4, -4) and B(-1, -3).

Equation of a tangent at a point of a circle with the center at the origin

The direction vector of the tangent at the point P₁ of a circle and the radius vector of P₁ are perpendicular to each other so their scalar product is zero. Points, O, P₁ and P in the right figure, determine vectors,

the scalar product written in the components gives,

x₁x + y₁y = r²

This is equation of a tangent at the point P₁(x₁, y₁) of a circle with the center at the origin.

Equation of a tangent at a point of a translated circle (x - p)² + (y - q)² = r²

The direction vector of the tangent at the point P₁(x₁, y₁), of a circle whose center is at the point S(p, q), and the direction vector of the normal, are perpendicular, so their scalar product is zero.