# Grade 10 Optional Mathematics: The Circle
## Unit: Coordinate Geometry | Chapter: Circle | Complete Solutions

---

## 🎯 Chapter Introduction

This guide offers a complete overview of the **Circle** chapter in Coordinate Geometry. It is designed to make the concepts easy to understand, with clear definitions, essential formulas, and detailed, step-by-step solutions to a wide range of problems. By the end of this note, you will have mastered the different forms of the circle equation and how to find a circle's center, radius, and equation based on various given conditions. ⚪

### Key Learning Areas
1.  **Standard Equation of a Circle:** Understanding and using the center-radius form.
2.  **General Equation of a Circle:** Identifying the center and radius from the general form.
3.  **Diameter Form:** Finding the equation of a circle given the endpoints of a diameter.
4.  **Problem Solving:** Finding the equation of a circle based on various conditions, such as passing through points, having a specific center, or being concentric with another circle.

---

## 🔑 Key Concepts & Formulas

Before we solve the problems, let's understand the fundamental equations of a circle.

### 1. Standard Equation of a Circle (Center-Radius Form)

This is the most common form of a circle's equation. It directly uses the circle's center and radius.

* **Formula:**
    $$(x - h)^2 + (y - k)^2 = r^2$$
    Where:
    * `(h, k)` are the coordinates of the center of the circle.
    * `r` is the radius of the circle.



---

### 2. General Equation of a Circle

This form is derived by expanding the standard equation. It is useful for identifying a circle's properties when the equation is not in center-radius form.

* **Formula:**
    $$x^2 + y^2 + 2gx + 2fy + c = 0$$

* **Finding the Center and Radius from the General Form:**
    * **Center:** $(-g, -f)$
    * **Radius:** $r = \sqrt{g^2 + f^2 - c}$

    To find `g`, `f`, and `c`, you compare the given equation to the general form. For example, if the equation is $x^2 + y^2 - 10x - 4y - 7 = 0$:
    * `2gx = -10x`  => `g = -5`
    * `2fy = -4y`  => `f = -2`
    * `c = -7`

---

### 3. Equation of a Circle in Diameter Form

If you know the coordinates of the two endpoints of a circle's diameter, you can use this special formula.

* **Formula:**
    $$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$
    Where $(x_1, y_1)$ and $(x_2, y_2)$ are the coordinates of the endpoints of a diameter.



---

### 4. Midpoint Formula

This is useful for finding the center of a circle if you know the endpoints of a diameter.

* **Formula:**
    $$Midpoint = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$$

---

## 📚 Exercise Solutions

### Short Questions

***

#### Question 1: Equation from Diameter Endpoints
**The endpoints of a diameter of a circle are (0, 6) and (8, 0). Find the equation of the circle.**

**Solution:**
Let $(x_1, y_1) = (0, 6)$ and $(x_2, y_2) = (8, 0)$.
Using the diameter form of the circle equation:
$$
\begin{aligned}
(x - x_1)(x - x_2) + (y - y_1)(y - y_2) &= 0 \\
(x - 0)(x - 8) + (y - 6)(y - 0) &= 0 \\
x(x - 8) + y(y - 6) &= 0 \\
x^2 - 8x + y^2 - 6y &= 0
\end{aligned}
$$
The required equation is **$x^2 + y^2 - 8x - 6y = 0$**.

***

#### Question 2: Equation from Diameter Endpoints
**The endpoints of a diameter of a circle are (3, 0) and (0, 4). Find the equation of the circle.**

**Solution:**
Let $(x_1, y_1) = (3, 0)$ and $(x_2, y_2) = (0, 4)$.
Using the diameter form:
$$
\begin{aligned}
(x - 3)(x - 0) + (y - 0)(y - 4) &= 0 \\
x(x - 3) + y(y - 4) &= 0 \\
x^2 - 3x + y^2 - 4y &= 0
\end{aligned}
$$
The required equation is **$x^2 + y^2 - 3x - 4y = 0$**.

***

#### Question 3: Equation from Diameter Endpoints
**Find the equation of the circle having (1, 3) and (2, -5) as the endpoints of a diameter.**

**Solution:**
Let $(x_1, y_1) = (1, 3)$ and $(x_2, y_2) = (2, -5)$.
Using the diameter form:
$$
\begin{aligned}
(x - 1)(x - 2) + (y - 3)(y - (-5)) &= 0 \\
(x^2 - 2x - x + 2) + (y - 3)(y + 5) &= 0 \\
x^2 - 3x + 2 + y^2 + 5y - 3y - 15 &= 0 \\
x^2 + y^2 - 3x + 2y - 13 &= 0
\end{aligned}
$$
The required equation is **$x^2 + y^2 - 3x + 2y - 13 = 0$**.

***

#### Question 4: Equation from Diameter Endpoints
**Find the equation of the circle having the endpoints of a diameter A(5, 6) and B(3, 4).**

**Solution:**
Let $(x_1, y_1) = (5, 6)$ and $(x_2, y_2) = (3, 4)$.
Using the diameter form:
$$
\begin{aligned}
(x - 5)(x - 3) + (y - 6)(y - 4) &= 0 \\
(x^2 - 3x - 5x + 15) + (y^2 - 4y - 6y + 24) &= 0 \\
x^2 - 8x + 15 + y^2 - 10y + 24 &= 0 \\
x^2 + y^2 - 8x - 10y + 39 &= 0
\end{aligned}
$$
The required equation is **$x^2 + y^2 - 8x - 10y + 39 = 0$**.

***

#### Question 5: Equation from Diameter Endpoints
**Find the equation of the circle whose endpoints of a diameter are A(3, 4) and B(2, -7).**

**Solution:**
Let $(x_1, y_1) = (3, 4)$ and $(x_2, y_2) = (2, -7)$.
Using the diameter form:
$$
\begin{aligned}
(x - 3)(x - 2) + (y - 4)(y - (-7)) &= 0 \\
(x^2 - 2x - 3x + 6) + (y - 4)(y + 7) &= 0 \\
x^2 - 5x + 6 + y^2 + 7y - 4y - 28 &= 0 \\
x^2 + y^2 - 5x + 3y - 22 &= 0
\end{aligned}
$$
The required equation is **$x^2 + y^2 - 5x + 3y - 22 = 0$**.

***

#### Question 6: Center from Intersecting Diameters
**Find the coordinates of the center of a circle having equations of two diameters $x + y = 5$ and $2x - y = 1$.**

**Solution:**
The center of a circle is the point of intersection of its diameters. We solve the system of linear equations.
$$
\begin{aligned}
x + y &= 5 \quad &(1) \\
2x - y &= 1 \quad &(2)
\end{aligned}
$$
Adding equation (1) and (2):
$$
\begin{aligned}
(x+y) + (2x-y) &= 5 + 1 \\
3x &= 6 \\
x &= 2
\end{aligned}
$$
Substitute $x=2$ into equation (1):
$$
2 + y = 5 \implies y = 3
$$
The center of the circle is **(2, 3)**.

***

#### Question 7: Center and Radius from General Equation
**Find the center and radius of a circle having the equation $x^2 + y^2 - 10x - 4y - 7 = 0$.**

**Solution:**
Compare the given equation to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.
$$
\begin{aligned}
2gx = -10x &\implies g = -5 \\
2fy = -4y &\implies f = -2 \\
c &= -7
\end{aligned}
$$
**Center:** $(-g, -f) = (-(-5), -(-2)) = (5, 2)$.
**Radius:** $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-5)^2 + (-2)^2 - (-7)} = \sqrt{25 + 4 + 7} = \sqrt{36} = 6$.

The center is **(5, 2)** and the radius is **6 units**.

***

#### Question 8: Finding an Unknown from Radius
**If the radius of the circle $x^2 + y^2 - 4x - 6y - k = 0$ is 4 units, find the value of k.**

**Solution:**
Compare the equation to $x^2 + y^2 + 2gx + 2fy + c = 0$.
$$
\begin{aligned}
2gx = -4x &\implies g = -2 \\
2fy = -6y &\implies f = -3 \\
c &= -k
\end{aligned}
$$
Using the radius formula, with $r=4$:
$$
\begin{aligned}
r &= \sqrt{g^2 + f^2 - c} \\
4 &= \sqrt{(-2)^2 + (-3)^2 - (-k)} \\
4 &= \sqrt{4 + 9 + k} \\
4 &= \sqrt{13 + k}
\end{aligned}
$$
Squaring both sides:
$$
16 = 13 + k \implies k = 3
$$
Thus, the value of **$k = 3$**.

***

#### Question 9: Finding Unknowns from Center
**If the center of the circle $x^2 + y^2 - ax - by - 12 = 0$ is (2, 3), find the value of a and b.**

**Solution:**
Compare the equation to $x^2 + y^2 + 2gx + 2fy + c = 0$.
$$
\begin{aligned}
2gx = -ax &\implies g = -a/2 \\
2fy = -by &\implies f = -b/2
\end{aligned}
$$
The center is given by $(-g, -f)$.
$$
Center = (-(-a/2), -(-b/2)) = (a/2, b/2)
$$
We are given that the center is (2, 3).
$$
\begin{aligned}
a/2 &= 2 \implies a = 4 \\
b/2 &= 3 \implies b = 6
\end{aligned}
$$
Thus, **$a=4$ and $b=6$**.

***

#### Question 10: Finding an Unknown from Center
**Find the value of m when the center of a circle with equation $x^2 + y^2 - mx + 6y - 12 = 0$ is (-2, -3).**

**Solution:**
Compare the equation to $x^2 + y^2 + 2gx + 2fy + c = 0$.
$$
\begin{aligned}
2gx = -mx &\implies g = -m/2 \\
2fy = 6y &\implies f = 3
\end{aligned}
$$
The center is $(-g, -f) = (-(-m/2), -3) = (m/2, -3)$.
We are given that the center is $(-2, -3)$.
Equating the x-coordinates:
$$
\frac{m}{2} = -2 \implies m = -4
$$
Thus, **$m = -4$**.

***

#### Question 11: Equation from Center and Radius
**Find the equation of a circle having centre (1, -2) and diameter $2\sqrt{5}$ units.**

**Solution:**
Center $(h, k) = (1, -2)$.
Diameter is $2\sqrt{5}$, so the radius is $r = \sqrt{5}$.
Using the standard form $(x-h)^2 + (y-k)^2 = r^2$:
$$
\begin{aligned}
(x - 1)^2 + (y - (-2))^2 &= (\sqrt{5})^2 \\
(x^2 - 2x + 1) + (y^2 + 4y + 4) &= 5 \\
x^2 + y^2 - 2x + 4y + 5 &= 5
\end{aligned}
$$
The required equation is **$x^2 + y^2 - 2x + 4y = 0$**.

***

#### Question 12: Equation from Center and Radius
**Find the equation of a circle having centre (2, 3) and radius 5 units.**

**Solution:**
Center $(h, k) = (2, 3)$ and radius $r = 5$.
Using the standard form:
$$
\begin{aligned}
(x - 2)^2 + (y - 3)^2 &= 5^2 \\
(x^2 - 4x + 4) + (y^2 - 6y + 9) &= 25 \\
x^2 + y^2 - 4x - 6y + 13 &= 25
\end{aligned}
$$
The required equation is **$x^2 + y^2 - 4x - 6y - 12 = 0$**.

***

#### Question 13: Center from General Equation
**Find the coordinates of the center of a circle having an equation $x^2 + y^2 + 4x - 5 = 0$.**

**Solution:**
Compare with $x^2 + y^2 + 2gx + 2fy + c = 0$.
$$
\begin{aligned}
2gx = 4x &\implies g = 2 \\
2fy = 0y &\implies f = 0
\end{aligned}
$$
The center is $(-g, -f) = (-2, 0)$.
The center is **(-2, 0)**.

***

#### Question 14: Equation from Center and Radius
**Find the equation of a circle having the centre (-1, -3) and radius 3 units.**

**Solution:**
Center $(h, k) = (-1, -3)$ and radius $r=3$.
Using the standard form:
$$
\begin{aligned}
(x - (-1))^2 + (y - (-3))^2 &= 3^2 \\
(x+1)^2 + (y+3)^2 &= 9 \\
(x^2 + 2x + 1) + (y^2 + 6y + 9) &= 9 \\
x^2 + y^2 + 2x + 6y + 1 &= 0
\end{aligned}
$$
The required equation is **$x^2 + y^2 + 2x + 6y + 1 = 0$**.

***

#### Question 15: Center from General Equation
**Find the center of the circle whose equation is $x^2 + y^2 - 6x + 2y + 1 = 0$.**

**Solution:**
Compare with $x^2 + y^2 + 2gx + 2fy + c = 0$.
$$
\begin{aligned}
2gx = -6x &\implies g = -3 \\
2fy = 2y &\implies f = 1
\end{aligned}
$$
The center is $(-g, -f) = (-(-3), -1) = (3, -1)$.
The center is **(3, -1)**.

---

### Long Questions

***

#### Question 16: Equation of Circle Passing Through Center of Another
**Circle A with centre X(-4, 5) passes through the centre Y of circle B. If the equation of circle B is $x^2 + y^2 - 4x + 6y - 12 = 0$, find the equation of circle A.**

**Solution:**
**Step 1: Find the center of circle B.**
Compare $x^2 + y^2 - 4x + 6y - 12 = 0$ with the general form.
$g = -2$, $f=3$.
Center of B is $Y = (-g, -f) = (2, -3)$.

**Step 2: Find the radius of circle A.**
The radius of circle A is the distance between its center X(-4, 5) and the point it passes through, Y(2, -3).
$$
\begin{aligned}
r^2 &= (x_2 - x_1)^2 + (y_2 - y_1)^2 \\
&= (2 - (-4))^2 + (-3 - 5)^2 \\
&= (6)^2 + (-8)^2 \\
&= 36 + 64 = 100
\end{aligned}
$$
**Step 3: Find the equation of circle A.**
The center of circle A is X(h, k) = (-4, 5) and $r^2=100$.
$$
\begin{aligned}
(x - (-4))^2 + (y - 5)^2 &= 100 \\
(x+4)^2 + (y-5)^2 &= 100 \\
x^2 + 8x + 16 + y^2 - 10y + 25 &= 100 \\
x^2 + y^2 + 8x - 10y - 59 &= 0
\end{aligned}
$$
The required equation is **$x^2 + y^2 + 8x - 10y - 59 = 0$**.

***

#### Question 17: Equation of Circle Passing Through Center of Another
**Find the equation of a circle with centre (-1, 2) and passing through the centre of the circle having equation $x^2 + y^2 - 6x - 10y - 2 = 0$.**

**Solution:**
**Step 1: Find the center of the given circle.**
Comparing $x^2 + y^2 - 6x - 10y - 2 = 0$ to the general form.
$g=-3$, $f=-5$.
The center is $(-g, -f) = (3, 5)$.

**Step 2: Find the radius of the required circle.**
The required circle has center $(h,k)=(-1,2)$ and passes through $(3,5)$. The radius is the distance between these two points.
$$
\begin{aligned}
r^2 &= (3 - (-1))^2 + (5 - 2)^2 \\
&= (4)^2 + (3)^2 \\
&= 16 + 9 = 25
\end{aligned}
$$
**Step 3: Find the equation of the required circle.**
Center is $(-1, 2)$ and $r^2=25$.
$$
\begin{aligned}
(x - (-1))^2 + (y - 2)^2 &= 25 \\
(x+1)^2 + (y-2)^2 &= 25 \\
x^2 + 2x + 1 + y^2 - 4y + 4 &= 25 \\
x^2 + y^2 + 2x - 4y - 20 &= 0
\end{aligned}
$$
The required equation is **$x^2 + y^2 + 2x - 4y - 20 = 0$**.

***

#### Question 18: Finding the Other End of a Diameter
**If one endpoint of a diameter of the circle $x^2 + y^2 + 4x - 6y + 8 = 0$ is (-4, 4), find the coordinates of the other endpoint.**

**Solution:**
**Step 1: Find the center of the circle.**
Comparing the equation to the general form: $g=2$, $f=-3$.
The center is $(-g, -f) = (-2, 3)$.

**Step 2: Use the midpoint formula.**
The center of a circle is the midpoint of its diameter. Let the other endpoint be $(x_2, y_2)$.
$$
\begin{aligned}
\text{Center} &= \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \\
(-2, 3) &= \left( \frac{-4+x_2}{2}, \frac{4+y_2}{2} \right)
\end{aligned}
$$
Equating coordinates:
$$
-2 = \frac{-4+x_2}{2} \implies -4 = -4+x_2 \implies x_2=0
$$
$$
3 = \frac{4+y_2}{2} \implies 6 = 4+y_2 \implies y_2=2
$$
The coordinates of the other end are **(0, 2)**.

***

#### Question 19: Equation of a Concentric Circle
**Find the equation of a circle concentric with the circle $2x^2 + 2y^2 + 4x - 2y + 1 = 0$ and passing through the point (4,-2).**

**Solution:**
**Step 1: Find the center of the given circle.**
First, divide the equation by 2 to get the standard general form:
$x^2 + y^2 + 2x - y + \frac{1}{2} = 0$.
Comparing with the general form, $g=1$ and $f=-1/2$.
The center is $(-g, -f) = (-1, 1/2)$.

**Step 2: Find the radius of the required circle.**
Since the required circle is concentric, its center is also $(-1, 1/2)$. It passes through $(4, -2)$. The radius is the distance between these points.
$$
\begin{aligned}
r^2 &= (4 - (-1))^2 + (-2 - \frac{1}{2})^2 \\
&= (5)^2 + (-\frac{5}{2})^2 \\
&= 25 + \frac{25}{4} = \frac{100+25}{4} = \frac{125}{4}
\end{aligned}
$$
**Step 3: Write the equation of the required circle.**
Center is $(-1, 1/2)$ and $r^2 = 125/4$.
$$
\begin{aligned}
(x - (-1))^2 + (y - \frac{1}{2})^2 &= \frac{125}{4} \\
(x+1)^2 + (y - \frac{1}{2})^2 &= \frac{125}{4} \\
x^2+2x+1 + y^2 - y + \frac{1}{4} &= \frac{125}{4}
\end{aligned}
$$
Multiply by 4 to clear the fraction:
$$
\begin{aligned}
4x^2 + 8x + 4 + 4y^2 - 4y + 1 &= 125 \\
4x^2 + 4y^2 + 8x - 4y - 120 &= 0 \\
x^2 + y^2 + 2x - y - 30 &= 0
\end{aligned}
$$
The required equation is **$x^2 + y^2 + 2x - y - 30 = 0$**.

***

#### Question 20: Equation of a Concentric Circle
**Find the equation of a circle passing through the point (4, 3) and concentric with the circle having equation $x^2 + y^2 + 6x - 8y - 11 = 0$.**

**Solution:**
**Step 1: Find the center of the given circle.**
Compare $x^2 + y^2 + 6x - 8y - 11 = 0$ with the general form.
$g=3$, $f=-4$. The center is $(-g,-f) = (-3, 4)$.

**Step 2: Find the radius of the required circle.**
Since the new circle is concentric, its center is also $(-3, 4)$. It passes through the point $(4, 3)$. The radius is the distance between these points.
$$
\begin{aligned}
r^2 &= (4 - (-3))^2 + (3 - 4)^2 \\
&= (7)^2 + (-1)^2 \\
&= 49 + 1 = 50
\end{aligned}
$$
**Step 3: Write the equation of the required circle.**
Center is $(-3, 4)$ and $r^2 = 50$.
$$
\begin{aligned}
(x - (-3))^2 + (y - 4)^2 &= 50 \\
(x+3)^2 + (y-4)^2 &= 50 \\
x^2+6x+9 + y^2-8y+16 &= 50 \\
x^2+y^2+6x-8y-25 &= 0
\end{aligned}
$$
The required equation is **$x^2 + y^2 + 6x - 8y - 25 = 0$**.

***

#### Question 21: Equation of a Circle from Two Points and Center on a Line
**Find the equation of the circle whose centre lies on the line $x - 4y = 1$ and which passes through the points (3, 7) and (5, 5).**

**Solution:**
Let the center be $(h,k)$. Since it lies on the line $x - 4y = 1$, we have $h - 4k = 1 \implies h = 4k+1$.
The distance from the center to any point on the circle is the radius. So, the squared distance from $(h,k)$ to $(3,7)$ is the same as to $(5,5)$.
$$
\begin{aligned}
(h-3)^2 + (k-7)^2 &= (h-5)^2 + (k-5)^2 \\
h^2 - 6h + 9 + k^2 - 14k + 49 &= h^2 - 10h + 25 + k^2 - 10k + 25 \\
-6h - 14k + 58 &= -10h - 10k + 50 \\
4h - 4k &= -8 \\
h - k &= -2
\end{aligned}
$$
Now substitute $h = 4k+1$ into this new equation:
$$
\begin{aligned}
(4k+1) - k &= -2 \\
3k &= -3 \\
k &= -1
\end{aligned}
$$
Now find h: $h = 4(-1)+1 = -3$. The center is $(-3,-1)$.
Find the radius squared, $r^2$, using point $(3,7)$:
$$
r^2 = (3 - (-3))^2 + (7 - (-1))^2 = 6^2 + 8^2 = 36+64=100
$$
The equation is $(x+3)^2 + (y+1)^2 = 100$, which expands to **$x^2 + y^2 + 6x + 2y - 90 = 0$**.

***

#### Question 22: Equation of a Circle from Three Points
**Find the equation of the circle which passes through the points P(5, 7), Q(6, 6) and R(2, -2).**

**Solution:**
Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$.
The squared distance from the center $(h,k)$ to each point is equal.
**From P and Q:**
$$
\begin{aligned}
(h-5)^2 + (k-7)^2 &= (h-6)^2 + (k-6)^2 \\
h^2-10h+25 + k^2-14k+49 &= h^2-12h+36 + k^2-12k+36 \\
-10h-14k+74 &= -12h-12k+72 \\
2h - 2k + 2 &= 0 \implies h - k + 1 = 0 \implies h = k-1 \quad \text{(Eq. 1)}
\end{aligned}
$$
**From Q and R:**
$$
\begin{aligned}
(h-6)^2 + (k-6)^2 &= (h-2)^2 + (k-(-2))^2 \\
h^2-12h+36 + k^2-12k+36 &= h^2-4h+4 + k^2+4k+4 \\
-12h-12k+72 &= -4h+4k+8 \\
64 &= 8h + 16k \\
8 &= h + 2k \quad \text{(Eq. 2)}
\end{aligned}
$$
**Solve for h and k:**
Substitute Eq. 1 into Eq. 2:
$8 = (k-1) + 2k \implies 9 = 3k \implies k=3$.
Then, $h = 3-1=2$. The center is $(h,k)=(2,3)$.

**Find $r^2$ using point P(5,7):**
$r^2 = (5-2)^2 + (7-3)^2 = 3^2 + 4^2 = 9+16=25$.

**Write the equation:**
$$(x-2)^2 + (y-3)^2 = 25$$
The required equation is **$x^2 + y^2 - 4x - 6y - 12 = 0$**.