# Conics - Circles

### Mind Warm Ups

Problem 1: what are conic sections or conics?

Ans: consider the double cone of figure 1. Some of the cross sections (plane slices) that can be obtained from the double cone are:
(a) Single point
(b) Single line
(c) Double lines
(d) Circle
(e) Parabola
(f) Ellipse
(g) Hyperbola
These shapes are called conic sections or conics. It will be shown later in the ellipse problems that the circle is a special case of an ellipse. The parabola, ellipse and hyperbola are called non-degenerate conics while the single point, single line and double lines are called degenerate conics

Problem 2: the equation of a circle in R2 is given as:
(x - a)2 + (y - b)2 = r2
Determine the coordinates of the center of the circle and the length of its radius.

Ans: center of circle is at (a,b); radius is r units.

Problem 3: consider the following curve in R2:
x2 + y2 - 8x -10y + 5 = 0.
Is this curve a circle? if yes, determine the coordinates of its center and the length of its radius.

Ans: reduce the equation of the curve to the equation of a circle as represented in problem 2 by completing the squares:
(x - 4)2 + (y - 5)2 -16 -25 + 5 = 0
therefore, (x - 4)2 + (y - 5)2 = 62
therefore, curve is a circle with center at (4,5) and radius of 6 units
In general, an equation of the form:
x2 + y2 + fx + gy + h = 0
represents a circle with
center (-f/2, -g/2) and radius √(f2/4 + g2/4 - h).
so long as f2/4 + g2/4 - h > 0.

Problem 4 : Determine the set of points (x,y) in R2 that satisfies the following equations:
(a) x2 + y2 + x/2 + y/2 + 1 = 0.
(b) x2 + y2 - 4x + 8y + 20 = 0.
(c) 2x2 + 2y2 - 8x -10y + 5 = 0.

Ans: (a) f2/4 + g2/4 - h < 0, therefore no points (x,y) satisfy the equation.
(b) f2/4 + g2/4 - h = 0, therefore the equation represents a single point.
(c) Divide equation by 2. Then f = -4, g = -5 and h = 5/2.
f2/4 + g2/4 - h > 0, therefore,
equation represents a circle with center at (-f/2, -g/2) = (2,5/2) and radius is √ (f2/4 + g2/4 - h) = √(41)/2.

Problem 5 : C1 and C2 are mutually orthogonal intersecting circles with the following equations:
C1: x2 + y2 - 4x - g1y + 7 = 0
C2: x2 + y2 + 2x -8y + 5 = 0
Determine the value of g1.

Ans: Two intersecting circles are mutually orthogonal if and only if:
f1f2 + g1g2 = 2(h1 + h2)
Therefore, -8 + 8g1 = 24
Therefore, g1 = 4 and -g1 = -4

Problem 6: C1 and C2 are circles that intersect at distinct points P and Q.
C1: x2 + y2 + f1x + g1y + h1 = 0
C2: x2 + y2 + f2x + g2y + h2 = 0
What is the general form of the equation of the line and any other circle through P and Q?

Ans: the general form of the equation is as follows:
x2 + y2 + f1x + g1y + h1 + k(x2 + y2 + f2x + g2y + h2) = 0 --- (1); for some number k.
If k ≠ -1, this equation is one of the circles. If k = -1, then it is the equation of the line.

Problem 7: consider circles C1, C2 and equation (1) of problem 6.
what value of k will reduce equation (1) to the equation of circle C2?
What does this value imply with respect to equation (1) of problem 6?

Ans: k = ∞. Divide equation (1) of problem 6 by k and let k ----> ∞. The implication is that C2 is excluded from the group of circles represented by equation (1) of problem 6.

Problem 8:These two circles intersect at distinct points P and Q:
C1: 2x2 + 2y2 - 6x + 8y - 2 = 0
C2: 2x2 + 2y2 + 5x - 6y + 3 = 0
Determine the equation of the circle that passes through (1,0) and the points P and Q.

Ans: this equation is of the form:
x2 + y2 - 3x + 4y - 1 + k(x2 + y2 + (5/2)x - 3y + 3/2) = 0
If it passes through (1,0), then 1 - 3 -1 + k(1 + 5/2 +3/2) = 0. Therefore, k = 3/5.
Therefore, the equation of the circle is:
x2 + y2 - 3x + 4y - 1 + 3/5(x2 + y2 + (5/2)x - 3y + 3/2) = 0
which simplifies to:
16x2 + 16y2 - 15x + 22y - 1 = 0.

Problem 9: Determine the equation of the line that passes through the points of intersection of circles C1 and C2 in problem 8.

Ans: k = -1 in this case. Therefore the equation is:
x2 + y2 - 3x + 4y - 1 -1(x2 + y2 + (5/2)x - 3y + 3/2) = 0
Simplifies to:
11x - 14y + 5 = 0.

Problem 10: consider circles C1 and C2:
C1: x2 + y2 + 4x + 4y + 8 = 0
C2: x2 + y2 + 6x + 2y + 4 = 0
(a) What points in R2 satisfy the result obtained when C2 is subtracted from C1?

(b) Now consider the following two circles:
C3: x2 + y2 = 1
C4: x2 + y2 = 4
What points in R2 satisfy the result obtained when C3 is subtracted from C4?

Ans: No help here :).

Peter Oye Sagay