Equation Of A Circle Answer Key

Introduction

The equation of a circle is one of the most fundamental tools in analytic geometry, allowing us to describe any circle on the Cartesian plane with just a few numbers. This article walks through the standard and general forms of the circle equation, explains how to derive each form, shows step‑by‑step solutions to typical problems, and provides a ready‑to‑use answer key for common exercise types. Whether you are preparing for a high‑school exam, tutoring a student, or reviewing a college‑level problem set, having a reliable answer key for circle equations can save time and boost confidence. By the end, you will be able to write, manipulate, and verify circle equations quickly and accurately.

1. Standard Form of a Circle

The most intuitive way to represent a circle is the standard form

[ (x-h)^2 + (y-k)^2 = r^2 ]

• ((h, k)) : coordinates of the center of the circle.
• (r) : radius (a non‑negative real number).

Why this form works

Starting from the definition of a circle—the set of all points that are a fixed distance (r) from a fixed point ((h,k))—we apply the distance formula:

[ \sqrt{(x-h)^2 + (y-k)^2}=r ;\Longrightarrow; (x-h)^2 + (y-k)^2 = r^2 . ]

Because the square root is eliminated, the equation becomes a polynomial, which is easier to handle algebraically.

Example 1: Write the equation of a circle with centre ((3,-2)) and radius (5).

[ (x-3)^2 + (y+2)^2 = 5^2 \quad\Longrightarrow\quad (x-3)^2 + (y+2)^2 = 25. ]

Answer key: ((x-3)^2 + (y+2)^2 = 25). Practical, not theoretical.

2. General Form of a Circle

Expanding the standard form yields the general (or expanded) form

[ x^2 + y^2 + Dx + Ey + F = 0, ]

where (D), (E), and (F) are constants related to the centre and radius:

[ h = -\frac{D}{2}, \qquad k = -\frac{E}{2}, \qquad r = \sqrt{h^2 + k^2 - F}. ]

Converting between forms

1. Standard → General – Expand the squares and move all terms to one side.
2. General → Standard – Complete the square for the (x)‑ and (y)‑terms.

Example 2: Convert (x^2 + y^2 - 6x + 8y + 9 = 0) to standard form.

1. Group (x) and (y) terms:

   [ (x^2 - 6x) + (y^2 + 8y) = -9. ]

2. Complete the square:

   [ (x^2 - 6x + 9) + (y^2 + 8y + 16) = -9 + 9 + 16. ]

3. Write as squares:

   [ (x-3)^2 + (y+4)^2 = 16. ]

Answer key: centre ((3,-4)), radius (4); equation ((x-3)^2 + (y+4)^2 = 16).

3. Common Problem Types and Solution Strategies

Below is a collection of typical exercise prompts you may encounter, each followed by a concise solution and the corresponding answer key entry.

3.1. Finding the Equation from Three Points

Problem: Determine the equation of the circle passing through the points (A(1,2)), (B(4,6)), and (C(-2,5)).

Strategy:

• Use the general form (x^2 + y^2 + Dx + Ey + F = 0).
• Substitute each point to generate three linear equations in (D, E, F).
• Solve the system (e.g., by elimination or matrix methods).

Solution:

[ \begin{cases} 1^2 + 2^2 + D(1) + E(2) + F = 0 \ 4^2 + 6^2 + D(4) + E(6) + F = 0 \ (-2)^2 + 5^2 + D(-2) + E(5) + F = 0 \end{cases} \Longrightarrow \begin{cases} 5 + D + 2E + F = 0 \ 52 + 4D + 6E + F = 0 \ 29 - 2D + 5E + F = 0 \end{cases} ]

Subtract the first equation from the second and third:

[ \begin{aligned} (52-5) + (4D-D) + (6E-2E) &= 0 ;\Rightarrow; 47 + 3D + 4E = 0,\ (29-5) + (-2D-D) + (5E-2E) &= 0 ;\Rightarrow; 24 - 3D + 3E = 0. \end{aligned} ]

Now solve the two‑equation system:

[ \begin{cases} 3D + 4E = -47\ -3D + 3E = -24 \end{cases} \Longrightarrow; D = -7,; E = -5. ]

Insert (D) and (E) into the first original equation:

[ 5 -7 + 2(-5) + F = 0 ;\Rightarrow; 5 -7 -10 + F = 0 ;\Rightarrow; F = 12. ]

Thus the general form is

[ x^2 + y^2 -7x -5y + 12 = 0. ]

Convert to standard form (complete the square):

[ (x^2 -7x + \tfrac{49}{4}) + (y^2 -5y + \tfrac{25}{4}) = -12 + \tfrac{49}{4} + \tfrac{25}{4} ] [ \Longrightarrow (x-\tfrac{7}{2})^2 + (y-\tfrac{5}{2})^2 = \frac{49+25-48}{4}= \frac{26}{4}= \frac{13}{2}. ]

Answer key:

• General form: (x^2 + y^2 -7x -5y + 12 = 0).
• Standard form: ((x-\tfrac{7}{2})^2 + (y-\tfrac{5}{2})^2 = \tfrac{13}{2}).
• Centre (\bigl(\tfrac{7}{2},\tfrac{5}{2}\bigr)), radius (\sqrt{\tfrac{13}{2}}).

3.2. Determining the Circle Tangent to a Line

Problem: Find the equation of the circle with centre ((2, -1)) that is tangent to the line (3x - 4y + 12 = 0).

Strategy:

• The distance from the centre to the line equals the radius (r).

• Use the point‑to‑line distance formula

  [ r = \frac{|3(2) - 4(-1) + 12|}{\sqrt{3^2 + (-4)^2}}. ]

Solution:

[ r = \frac{|6 + 4 + 12|}{\sqrt{9 + 16}} = \frac{22}{5}. ]

Insert (r) into the standard equation:

[ (x-2)^2 + (y+1)^2 = \left(\frac{22}{5}\right)^2 = \frac{484}{25}. ]

Answer key: ((x-2)^2 + (y+1)^2 = \frac{484}{25}).

3.3. Circle Passing Through the Origin and Two Given Points

Problem: Write the equation of the circle that passes through the origin ((0,0)) and the points (P(4,0)) and (Q(0,3)).

Strategy:

• Since the origin lies on the circle, the constant term (F) in the general form is zero.
• Use the remaining two points to solve for (D) and (E).

Solution:

General form: (x^2 + y^2 + Dx + Ey = 0).

Plug (P(4,0)): (16 + 0 + 4D + 0 = 0 ;\Rightarrow; 4D = -16 ;\Rightarrow; D = -4.)

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Plug (Q(0,3)): (0 + 9 + 0 + 3E = 0 ;\Rightarrow; 3E = -9 ;\Rightarrow; E = -3.)

Thus the equation is

[ x^2 + y^2 -4x -3y = 0. ]

Complete the square:

[ (x^2 -4x + 4) + (y^2 -3y + \tfrac{9}{4}) = 4 + \tfrac{9}{4} ] [ \Longrightarrow (x-2)^2 + \left(y-\tfrac{3}{2}\right)^2 = \frac{25}{4}. ]

Answer key:

• General form: (x^2 + y^2 -4x -3y = 0).
• Standard form: ((x-2)^2 + (y-\tfrac{3}{2})^2 = \tfrac{25}{4}).
• Centre ((2,\tfrac{3}{2})), radius (\tfrac{5}{2}).

3.4. Finding the Equation of a Circle Orthogonal to Two Given Circles

Problem: Determine the equation of a circle orthogonal to both

[ C_1:;(x-1)^2 + (y+2)^2 = 9,\qquad C_2:;(x+3)^2 + (y-1)^2 = 16, ]

and whose centre lies on the line (y = x).

Strategy:

• For orthogonal circles, the relationship between centres ((h_1,k_1)), ((h_2,k_2)) and radii (r_1, r_2) of the unknown circle ((h,k,r)) is

  [ (h-h_1)^2 + (k-k_1)^2 = r^2 + r_1^2, ] [ (h-h_2)^2 + (k-k_2)^2 = r^2 + r_2^2. ]

• Because the centre lies on (y = x), set (k = h).

• Solve the resulting two equations for (h) and (r).

Solution:

Let ((h,h)) be the centre, (r) the radius.

From orthogonality with (C_1) ((h_1=1, k_1=-2, r_1=3)):

[ (h-1)^2 + (h+2)^2 = r^2 + 9 \quad\Longrightarrow\quad 2h^2 + 2h + 5 = r^2 + 9. \tag{1} ]

From orthogonality with (C_2) ((h_2=-3, k_2=1, r_2=4)):

[ (h+3)^2 + (h-1)^2 = r^2 + 16 \quad\Longrightarrow\quad 2h^2 + 4h + 10 = r^2 + 16. \tag{2} ]

Subtract (1) from (2):

[ (2h^2 + 4h + 10) - (2h^2 + 2h + 5) = (r^2 + 16) - (r^2 + 9) ] [ 2h + 5 = 7 ;\Longrightarrow; h = 1. ]

Thus the centre is ((1,1)). Plug back into (1):

[ 2(1)^2 + 2(1) + 5 = r^2 + 9 ;\Longrightarrow; 2 + 2 + 5 = r^2 + 9 ;\Longrightarrow; 9 = r^2 + 9 ;\Longrightarrow; r^2 = 0. ]

A radius of zero indicates the only circle orthogonal to both and whose centre lies on (y=x) is the point circle at ((1,1)). If a non‑degenerate circle is required, the condition “orthogonal to both” cannot be satisfied simultaneously with the centre on (y=x).

Answer key: Centre ((1,1)), radius (0) (point circle).

(This example illustrates how the answer key can also reveal when a problem has no non‑trivial solution.)

4. Frequently Asked Questions

Q1: Can a circle have a negative radius in its equation?

A: No. The radius (r) is defined as a non‑negative distance. If you obtain (r^2 < 0) after completing the square, the original equation does not represent a real circle; it may represent an empty set or a complex‑valued curve.

Q2: What does it mean when the coefficient of (x^2) and (y^2) are not both 1?

A: For a true circle, the coefficients of (x^2) and (y^2) must be equal and non‑zero. If they differ, the curve is an ellipse (or a hyperbola if one coefficient is negative). Scaling both sides by the same constant can sometimes restore the coefficient to 1, but only if the coefficients are already equal.

Q3: How can I quickly check whether a given quadratic equation represents a circle?

A: Verify three conditions:

1. Coefficients of (x^2) and (y^2) are equal and non‑zero.
2. No (xy) term is present.
3. After completing the square, the right‑hand side (the squared radius) is positive.

If all hold, the equation describes a real circle.

Q4: Is the “answer key” for circle equations the same as a solution manual?

A: In educational contexts, an answer key typically lists the final equations, centre coordinates, and radii without showing the intermediate algebra. A solution manual provides step‑by‑step reasoning. The answer key presented here includes both concise results and brief derivations for clarity.

Q5: Why do some textbooks use the form ((x-h)^2 + (y-k)^2 = r^2) while others write (x^2 + y^2 + Dx + Ey + F = 0)?

A: The standard form highlights geometric intuition (centre & radius). The general form is convenient for algebraic manipulation, especially when working with multiple circles simultaneously (e.g., finding intersections, orthogonal circles). Mastery of both forms is essential for flexibility.

5. Practical Tips for Working Efficiently with Circle Equations

• Always start by identifying the given data (centre, radius, a point on the circle, tangency condition, etc.). Write them down in a list before manipulating equations.
• When converting to general form, expand systematically: ((x-h)^2 = x^2 - 2hx + h^2). Do the same for the (y) part, then combine like terms.
• Complete the square in one go: group all (x) terms together, factor the coefficient of (x^2) if it isn’t 1, then add/subtract the needed constant on both sides.
• Use the distance formula for tangency or orthogonal problems; it often reduces a messy algebraic condition to a simple numeric radius.
• Check your final answer by plugging a known point (e.g., the centre or a point given in the problem) back into the derived equation. If it satisfies the equation, you have likely avoided arithmetic errors.

6. Complete Answer Key Summary

Problem Type	Final Equation (Standard)	Final Equation (General)	Centre ((h,k))	Radius (r)
Centre ((3,-2)), (r=5)	((x-3)^2 + (y+2)^2 = 25)	(x^2 + y^2 -6x +4y + 0 = 0)	((3,-2))	5
Convert (x^2 + y^2 - 6x + 8y + 9 = 0)	((x-3)^2 + (y+4)^2 = 16)	—	((3,-4))	4
Circle through (A(1,2), B(4,6), C(-2,5))	((x-\tfrac{7}{2})^2 + (y-\tfrac{5}{2})^2 = \tfrac{13}{2})	(x^2 + y^2 -7x -5y + 12 = 0)	((\tfrac{7}{2},\tfrac{5}{2}))	(\sqrt{\tfrac{13}{2}})
Centre ((2,-1)) tangent to (3x-4y+12=0)	((x-2)^2 + (y+1)^2 = \frac{484}{25})	—	((2,-1))	(\frac{22}{5})
Through ((0,0),(4,0),(0,3))	((x-2)^2 + (y-\tfrac{3}{2})^2 = \tfrac{25}{4})	(x^2 + y^2 -4x -3y = 0)	((2,\tfrac{3}{2}))	(\tfrac{5}{2})
Orthogonal to (C_1, C_2) with centre on (y=x)	Point circle ((x-1)^2 + (y-1)^2 = 0)	(x^2 + y^2 -2x -2y + 2 = 0)	((1,1))	0

Conclusion

Mastering the equation of a circle equips you with a versatile tool for geometry, calculus, physics, and computer graphics. Still, by internalizing the transition between standard and general forms, practicing the typical problem patterns presented above, and referring to the concise answer key, you can solve circle‑related tasks quickly and with confidence. In practice, remember to verify each step, keep the geometric meaning of the parameters in mind, and use the distance formula whenever tangency or orthogonality appears. With these strategies, the circle equation becomes not just a formula to memorize, but a reliable companion for any analytic‑geometry challenge.