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Equations and Properties of Circles

CSCA Equations and Properties of Circles study guide organized around the publicly available CSCA syllabus. Practice Mathematics questions on aicsca.com.

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Equations and Properties of Circles

Circles are a core topic in the CSCA geometry section, often tested in combination with lines and distance formulas. Mastering the two forms of circle equations and the "distance from center to line" is key to solving these problems.

1. Definition and Standard Equation

A circle is the set of all points in a plane that are at a fixed distance (Radius) from a fixed point (Center).

Let the center be $C(a, b)$ and the radius be $r$ $(r > 0)$. Using the distance formula, the **Standard Equation** is:

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

* **Special Case**: If the center is at the origin $O(0,0)$, the equation simplifies to $x^2 + y^2 = r^2$.

2. General Equation

Expanding and rearranging the standard equation gives the **General Equation**:

$$x^2 + y^2 + Dx + Ey + F = 0$$

* **Center**: $(-\frac{D}{2}, -\frac{E}{2})$

* **Radius**: $r = \frac{1}{2}\sqrt{D^2 + E^2 - 4F}$

* **Condition**: This equation represents a real circle only if $D^2 + E^2 - 4F > 0$.

3. Positional Relationships

This is the most frequently tested topic, especially the relationship between a "Line and a Circle".

#### 3.1 Point and Circle

Calculate the distance $d$ from point $P$ to center $C$:

* $d > r \Leftrightarrow$ Outside the circle

* $d = r \Leftrightarrow$ On the circle

* $d < r \Leftrightarrow$ Inside the circle

#### 3.2 Line and Circle (**Important**)

To determine the relationship between a line $l: Ax + By + C = 0$ and a circle, the **Geometric Method** is preferred: compare the distance $d$ (from center to line) with the radius $r$.

Distance formula from center $(a, b)$ to line $l$:

$$d = \frac{|Aa + Bb + C|}{\sqrt{A^2 + B^2}}$$

* **Intersecting**: $d < r$ $\Rightarrow$ Two intersection points (Secant line).

* **Tangent**: $d = r$ $\Rightarrow$ One intersection point (Tangent line). **This is the core condition for finding tangent equations.**

* **Disjoint**: $d > r$ $\Rightarrow$ No intersection points.

4. Equation of Tangent Line

1. **Tangent at a Point on the Circle** (Speed Trick):

If point $P(x_0, y_0)$ is **on** the circle $x^2 + y^2 = r^2$, the tangent line is:

$$x_0x + y_0y = r^2$$

For the general center $(x-a)^2 + (y-b)^2 = r^2$, it is:

$$(x_0-a)(x-a) + (y_0-b)(y-b) = r^2$$

2. **Tangent from an External Point**:

Assume the slope is $k$, write the point-slope form equation, and solve for $k$ using the condition "distance from center to line $d=r$".

5. Practice Examples

**Example 1**: Find the equation of the circle with center $C(1, -2)$ passing through $P(4, 2)$.

**Solution**:

Radius $r = |CP| = \sqrt{(4-1)^2 + (2-(-2))^2} = \sqrt{9+16} = 5$.

Substitute into standard equation: $(x-1)^2 + (y+2)^2 = 25$.

**Example 2**: Determine the relationship between line $3x - 4y + 15 = 0$ and circle $x^2 + y^2 = 9$.

**Solution**:

Center is $(0,0)$, radius $r=3$.

Distance from center to line $d = \frac{|3(0) - 4(0) + 15|}{\sqrt{3^2 + (-4)^2}} = \frac{15}{5} = 3$.

Since $d = r$, the line is **Tangent** to the circle.

**Example 3**: Find the center and radius of $x^2 + y^2 - 4x + 6y - 12 = 0$.

**Solution**:

* Method 1 (Formula): $D=-4, E=6, F=-12$.

Center $(2, -3)$, Radius $r = \frac{1}{2}\sqrt{16 + 36 - 4(-12)} = 5$.

* Method 2 (Completing the Square):

$(x-2)^2 + (y+3)^2 = 25$.

Center $(2, -3)$, Radius $5$.