Homework 8 Equations of Circles Answers – A Complete Guide
When tackling homework 8 equations of circles, students often face two main challenges: correctly identifying the form of the circle’s equation and applying the right algebraic techniques to solve for the unknowns. This guide not only provides the answers to the typical problems found in such a homework set but also walks through the reasoning behind each step. By the end, you’ll understand the underlying geometry and be equipped to solve any circle‑equation problem confidently.
Introduction
A circle in the Cartesian plane can be described by several equivalent equations, each useful in different contexts:
-
Standard form
[(x - h)^2 + (y - k)^2 = r^2]
where ((h, k)) is the center and (r) is the radius. -
General form
[x^2 + y^2 + Dx + Ey + F = 0]
obtained by expanding the standard form and collecting like terms. -
Parametric form (rarely used in basic algebra)
[x = h + r\cos\theta,\quad y = k + r\sin\theta.]
Homework problems usually present one of these forms or a mix of them, requiring you to:
- Identify the missing parameters (center, radius, or constants).
- Verify whether a given point lies on the circle.
- Find the circle that passes through three given points.
- Convert between standard and general forms.
Below, we provide detailed solutions for eight representative problems, illustrating each common type.
Problem 1 – Find the Center and Radius
Question
Given the equation
[(x - 3)^2 + (y + 5)^2 = 16,]
determine the center and radius of the circle.
Solution
The equation is already in standard form ((x - h)^2 + (y - k)^2 = r^2).
- Center ((h, k) = (3, -5)).
- Radius (r = \sqrt{16} = 4).
Answer
Center: ((3, -5)); Radius: (4).
Problem 2 – Expand to General Form
Question
Expand the standard equation [(x + 2)^2 + (y - 4)^2 = 25] into general form.
Solution
Expand each squared term:
[ (x + 2)^2 = x^2 + 4x + 4, \quad (y - 4)^2 = y^2 - 8y + 16. ]
Add them and set equal to 25:
[ x^2 + 4x + 4 + y^2 - 8y + 16 = 25. ]
Combine constants:
[ x^2 + y^2 + 4x - 8y + 20 - 25 = 0 ;\Rightarrow; x^2 + y^2 + 4x - 8y - 5 = 0. ]
Answer
General form: (\boxed{x^2 + y^2 + 4x - 8y - 5 = 0}) Surprisingly effective..
Problem 3 – Verify a Point Lies on the Circle
Question
Does the point ((1, 3)) lie on the circle [x^2 + y^2 - 6x + 8y + 9 = 0?]
Solution
Substitute (x = 1), (y = 3):
[ 1^2 + 3^2 - 6(1) + 8(3) + 9 = 1 + 9 - 6 + 24 + 9 = 37. ]
Since (37 \neq 0), the point is not on the circle.
Answer
No, ((1, 3)) is not on the circle Worth keeping that in mind..
Problem 4 – Find the Circle Through Three Points
Question
Determine the equation of the circle passing through the points (A(1, 2)), (B(4, 6)), and (C(7, 2)).
Solution
For three non‑collinear points, the circle’s center ((h, k)) is the intersection of the perpendicular bisectors of any two chords.
-
Midpoint of AB:
(\big(\frac{1+4}{2}, \frac{2+6}{2}\big) = (2.5, 4)).
Slope of AB: (\frac{6-2}{4-1} = \frac{4}{3}).
Perpendicular slope: (-\frac{3}{4}).
Equation: ((y - 4) = -\frac{3}{4}(x - 2.5)). -
Midpoint of AC:
(\big(\frac{1+7}{2}, \frac{2+2}{2}\big) = (4, 2)).
Slope of AC: (\frac{2-2}{7-1} = 0).
Perpendicular slope: undefined (vertical line).
Equation: (x = 4) Practical, not theoretical..
Intersecting (x = 4) with the first bisector:
[ y - 4 = -\frac{3}{4}(4 - 2.5) = -\frac{3}{4}\cdot1.5 = -1.125 ;\Rightarrow; y = 2.875.
So center ((h, k) = (4, 2.875)).
Compute radius (r) using distance to point (A(1, 2)):
[ r = \sqrt{(4-1)^2 + (2.In practice, 875^2} = \sqrt{9 + 0. Think about it: 765625} = \sqrt{9. Still, 765625} \approx 3. Worth adding: 875-2)^2} = \sqrt{3^2 + 0. 125.
Thus the circle’s standard equation:
[ (x - 4)^2 + (y - 2.Still, 875)^2 = (3. 125)^2.
Simplify the radius squared:
[ (3.125)^2 = 9.765625. ]
Answer
[\boxed{(x - 4)^2 + (y - 2.875)^2 = 9.765625}]
(or any equivalent form).
Problem 5 – Convert General to Standard Form
Question
Rewrite the equation (x^2 + y^2 - 10x + 14y + 25 = 0) in standard form.
Solution
Group (x) and (y) terms, then complete the square Which is the point..
-
(x)-terms: (x^2 - 10x).
Half of (-10) is (-5); square gives (25).
Add and subtract (25). -
(y)-terms: (y^2 + 14y).
Half of (14) is (7); square gives (49).
Add and subtract (49) Simple as that..
Rewrite:
[ (x^2 - 10x + 25) + (y^2 + 14y + 49) + 25 + 49 + 25 = 0. ]
Simplify constants:
[ (x - 5)^2 + (y + 7)^2 + 99 = 0. ]
Move constant to the right:
[ (x - 5)^2 + (y + 7)^2 = -99. ]
Since the right side is negative, there is no real circle satisfying this equation (the set of points is empty). This is a trick question often included to test algebraic manipulation.
Answer
No real circle exists; the equation has no real solution Easy to understand, harder to ignore..
Problem 6 – Tangent Line to a Circle
Question
Find the equation of the tangent line to the circle ((x - 2)^2 + (y + 3)^2 = 16) at the point ((6, -3)) Less friction, more output..
Solution
The radius at the point of tangency is perpendicular to the tangent.
Radius vector: (\langle 6-2, -3-(-3)\rangle = \langle 4, 0\rangle).
Thus the slope of the radius is (0) (horizontal).
The tangent line must be vertical, so its equation is:
[ x = 6. ]
Answer
[\boxed{x = 6}]
Problem 7 – Intersection of Two Circles
Question
Determine the intersection points of the circles
[
\begin{cases}
(x - 1)^2 + (y + 2)^2 = 9,\
(x + 2)^2 + (y - 1)^2 = 4.
\end{cases}
]
Solution
Expand both equations and subtract to eliminate squares.
-
First circle: ((x - 1)^2 + (y + 2)^2 = 9).
Expand: (x^2 - 2x + 1 + y^2 + 4y + 4 = 9) → (x^2 + y^2 - 2x + 4y - 4 = 0). -
Second circle: ((x + 2)^2 + (y - 1)^2 = 4).
Expand: (x^2 + 4x + 4 + y^2 - 2y + 1 = 4) → (x^2 + y^2 + 4x - 2y + 1 = 0).
Subtract the first from the second:
[ (4x - 2y + 1) - (-2x + 4y - 4) = 0 ;\Rightarrow; 6x - 6y + 5 = 0 ;\Rightarrow; y = x + \frac{5}{6}. ]
Plug (y = x + \frac{5}{6}) into the first circle’s expanded form:
[ x^2 + (x + \tfrac{5}{6})^2 - 2x + 4(x + \tfrac{5}{6}) - 4 = 0. ]
Compute:
[ x^2 + x^2 + \tfrac{5}{3}x + \tfrac{25}{36} - 2x + 4x + \tfrac{20}{6} - 4 = 0. ]
Simplify step by step:
- Combine (x^2) terms: (2x^2).
- Combine (x) terms: (\tfrac{5}{3}x - 2x + 4x = \tfrac{5}{3}x + 2x = \tfrac{11}{3}x).
- Constants: (\tfrac{25}{36} + \tfrac{20}{6} - 4 = \tfrac{25}{36} + \tfrac{120}{36} - \tfrac{144}{36} = \tfrac{1}{36}).
Equation:
[ 2x^2 + \tfrac{11}{3}x + \tfrac{1}{36} = 0. ]
Multiply by 36 to clear denominators:
[ 72x^2 + 132x + 1 = 0. ]
Solve using quadratic formula:
[ x = \frac{-132 \pm \sqrt{132^2 - 4 \cdot 72 \cdot 1}}{2 \cdot 72} = \frac{-132 \pm \sqrt{17424 - 288}}{144} = \frac{-132 \pm \sqrt{17136}}{144} = \frac{-132 \pm 131}{144}. ]
Two solutions:
-
(x = \frac{-132 + 131}{144} = \frac{-1}{144} \approx -0.00694).
Then (y = x + \tfrac{5}{6} \approx -0.00694 + 0.83333 = 0.82639) Took long enough.. -
(x = \frac{-132 - 131}{144} = \frac{-263}{144} \approx -1.82639).
Then (y = x + \tfrac{5}{6} \approx -1.82639 + 0.83333 = -0.99306).
Answer
Intersection points:
[
\boxed{(-0.00694,; 0.82639)} \quad \text{and} \quad \boxed{(-1.82639,; -0.99306)}.
]
Problem 8 – Radius from an Equation with a Constant Term
Question
Find the radius of the circle described by (x^2 + y^2 + 8x - 6y + 20 = 0).
Solution
Rewrite in standard form by completing the square.
-
Group (x) terms: (x^2 + 8x). Half of 8 is 4; square gives 16.
Add and subtract 16. -
Group (y) terms: (y^2 - 6y). Half of (-6) is (-3); square gives 9.
Add and subtract 9.
Rewrite:
[ (x^2 + 8x + 16) + (y^2 - 6y + 9) + 20 - 16 - 9 = 0. ]
Simplify constants:
[ (x + 4)^2 + (y - 3)^2 - 5 = 0 ;\Rightarrow; (x + 4)^2 + (y - 3)^2 = 5. ]
Thus radius (r = \sqrt{5}) Worth keeping that in mind..
Answer
Radius: (\boxed{\sqrt{5}}) Most people skip this — try not to..
FAQ – Common Mistakes and Tips
| Question | Answer |
|---|---|
| *What if the constant on the right side of the standard form is negative?Now, * | That indicates there is no real circle; the set of points is empty. And |
| *How do I check if three points are collinear before solving for a circle? * | Compute the slope between two pairs. If both slopes are equal (or both vertical), the points are collinear and no unique circle exists. |
| *Why does completing the square sometimes introduce large numbers?And * | It’s a systematic procedure; the numbers are necessary to isolate ((x-h)^2) and ((y-k)^2). Keep track of additions and subtractions carefully. |
| Can a circle have a radius of zero? | Yes, that would be a point circle—the equation reduces to ((x-h)^2 + (y-k)^2 = 0). |
Conclusion
Mastering the eight standard equations of circles equips you to solve a vast array of problems—from finding centers and radii to determining tangents and intersections. The key steps are:
- Recognize the form (standard vs. general).
- Apply algebraic techniques (completing the square, expanding, subtracting).
- Verify solutions by substitution or geometric reasoning.
With practice, these problems become routine, and you’ll be able to approach more advanced topics such as circle families, orthogonal circles, and circle inversion with confidence. Happy solving!
The analysis concludes with the identified intersection points and verified radius, encapsulating the solution's closure.
\boxed{(-0.00694, 0.82639)}
and
\boxed{(-1.82639, -0.99306)}.
This summary underscores the resolution's efficacy.
