**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 R^{2} is given as:

(x - a)^{2} + (y - b)^{2} = r^{2}

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 R^{2}:

x^{2} + y^{2} - 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} = 6^{2}

therefore, curve is a circle with center at (4,5) and radius of 6 units

In general, an equation of the form:

x^{2} + y^{2} + fx + gy + h = 0

represents a circle with

center (-f/2, -g/2) and radius √(f^{2}/4 + g^{2}/4 - h).

so long as f^{2}/4 + g^{2}/4 - h > 0.

**Problem 4 **: Determine the set of points (x,y) in R^{2} that satisfies the following equations:

(a) x^{2} + y^{2} + x/2 + y/2 + 1 = 0.

(b) x^{2} + y^{2} - 4x + 8y + 20 = 0.

(c) 2x^{2} + 2y^{2} - 8x -10y + 5 = 0.
**Ans: **(a) f^{2}/4 + g^{2}/4 - h < 0, therefore no points (x,y) satisfy the equation.

(b) f^{2}/4 + g^{2}/4 - h = 0, therefore the equation represents a single point.

(c) Divide equation by 2. Then f = -4, g = -5 and h = 5/2.

f^{2}/4 + g^{2}/4 - h > 0, therefore,

equation represents a circle with center at (-f/2, -g/2) = (2,5/2) and radius is √ (f^{2}/4 + g^{2}/4 - h) = √(41)/2.

**Problem 5 **: C_{1} and C_{2} are mutually orthogonal intersecting circles with the following equations:

C_{1}: x^{2} + y^{2} - 4x - g_{1}y + 7 = 0

C_{2}: x^{2} + y^{2} + 2x -8y + 5 = 0

Determine the value of g_{1}.
**Ans:**
Two intersecting circles are mutually orthogonal if and only if:

f_{1}f_{2} + g_{1}g_{2} = 2(h_{1} + h_{2})

Therefore, -8 + 8g_{1} = 24

Therefore, g_{1} = 4 and -g_{1} = -4

**Problem 6**: C_{1} and C_{2} are circles that intersect at distinct points P and Q.

C_{1}: x^{2} + y^{2} + f_{1}x + g_{1}y + h_{1} = 0

C_{2}: x^{2} + y^{2} + f_{2}x + g_{2}y + h_{2} = 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:

x^{2} + y^{2} + f_{1}x + g_{1}y + h_{1} + k(x^{2} + y^{2} + f_{2}x + g_{2}y + h_{2}) = 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 C_{1}, C_{2} and equation (1) of problem 6.

what value of k will reduce equation (1) to the equation of circle C_{2}?

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 C_{2} 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:

C_{1}: 2x^{2} + 2y^{2} - 6x + 8y - 2 = 0

C_{2}: 2x^{2} + 2y^{2} + 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:

x^{2} + y^{2} - 3x + 4y - 1 + k(x^{2} + y^{2} + (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:

x^{2} + y^{2} - 3x + 4y - 1 + 3/5(x^{2} + y^{2} + (5/2)x - 3y + 3/2) = 0

which simplifies to:

16x^{2} + 16y^{2} - 15x + 22y - 1 = 0.

**Problem 9**: Determine the equation of the line that passes through the points of intersection of circles C_{1} and C_{2} in problem 8.
**Ans: **k = -1 in this case. Therefore the equation is:

x^{2} + y^{2} - 3x + 4y - 1 -1(x^{2} + y^{2} + (5/2)x - 3y + 3/2) = 0

Simplifies to:

11x - 14y + 5 = 0.

**Problem 10**: consider circles C_{1} and C_{2}:

C_{1}: x^{2} + y^{2} + 4x + 4y + 8 = 0

C_{2}: x^{2} + y^{2} + 6x + 2y + 4 = 0

(a) What points in R^{2} satisfy the result obtained when C_{2} is subtracted from C_{1}?

(b) Now consider the following two circles:

C_{3}: x^{2} + y^{2} = 1

C_{4}: x^{2} + y^{2} = 4

What points in R^{2} satisfy the result obtained when C_{3} is subtracted from C_{4}?
**Ans: ** No help here :).

Peter Oye Sagay