Equation of a Circle and Tangents

The Equation of a Circle

For any circle centred at the origin $(0,0)$, the equation is derived directly from Pythagoras’ Theorem: for any point $(x,y)$ on the circumference, the radius $r$ is the hypotenuse.

Here $(x, y)$ is any point on the circumference and $r$ is the radius.

Tangents to a Circle

A tangent is a straight line that touches a circle at exactly one point. The crucial rule is:

This means the radius and tangent gradients are negative reciprocals of each other: $m_{\text{radius}} \times m_{\text{tangent}} = -1$.

Example 1: Finding the Equation of a Circle

A circle is centred at the origin and passes through $(6, 8)$. Find its equation.

1. Start with $x^2 + y^2 = r^2$.
2. Substitute $(6, 8)$: $6^2 + 8^2 = r^2$.
3. Calculate: $36 + 64 = 100$, so $r^2 = 100$.

Answer: $x^2 + y^2 = 100$

Example 2: Finding the Equation of a Tangent

Find the tangent to $x^2 + y^2 = 50$ at $(5, -5)$.

1. Gradient of radius: $m_r = \dfrac{-5 – 0}{5 – 0} = -1$.
2. Gradient of tangent (negative reciprocal): $m_t = -\dfrac{1}{-1} = 1$.
3. Line equation through $(5, -5)$ with $m=1$:
   $y – (-5) = 1(x – 5) \implies y = x – 10$.

Answer: $y = x – 10$

🤔 Common Misunderstandings

• Confusing $r$ and $r^2$. If the equation is $x^2 + y^2 = 25$, the radius is $\sqrt{25} = 5$, not 25.
• Gradient sign errors. Forgetting to negate when finding the negative reciprocal (e.g., $-\frac{1}{3}$ becomes $+3$, not $-3$).
• Wrong point in the line equation. Use the point on the circle — not the origin — when applying $y – y_1 = m(x – x_1)$.

📝 Common Exam Mistakes

• Not showing working for the perpendicular gradient — both the radius and tangent gradients must be shown clearly.
• Arithmetic errors with negative coordinates, especially when squaring negatives.
• Leaving the tangent unsimplified — check whether the question requires $y = mx + c$ form.