Equation of a Circle [H]

Equation of a Circle

A circle with centre at the origin \( (0, 0) \) and radius \( r \) has equation:

\[ x^2 + y^2 = r^2 \]

This follows directly from Pythagoras: any point \( (x, y) \) on the circle is a distance \( r \) from the origin, so \( \sqrt{x^2+y^2} = r \), giving \( x^2+y^2=r^2 \).

x y 2 −2 2 −2 r = 5 P(3, 4) Tangent O x² + y² = 25

Tangents to a Circle

A tangent to a circle is a straight line that touches the circle at exactly one point. The key geometric property:

The tangent at any point on a circle is perpendicular to the radius at that point.

This gives the method for finding a tangent equation:

  1. Find the gradient of the radius from the origin \(O\) to the point \(P(x_1, y_1)\): \( m_{\text{radius}} = \dfrac{y_1}{x_1} \).
  2. The gradient of the tangent is the negative reciprocal: \( m_{\text{tangent}} = -\dfrac{x_1}{y_1} \).
  3. Use point–slope form: \( y - y_1 = m_{\text{tangent}}(x - x_1) \).

Worked Examples

Write the equation of a circle with centre \(O\) and radius 7. State the radius of the circle \( x^2 + y^2 = 50 \).
\[ x^2 + y^2 = 49 \]

For \( x^2+y^2=50 \): \( r^2 = 50 \Rightarrow r = \sqrt{50} = 5\sqrt{2} \).

Find the equation of the tangent to \( x^2+y^2=25 \) at the point \( P(3, 4) \).

Gradient of radius \(OP\): \( m = 4/3 \).

Gradient of tangent: \( m_T = -3/4 \).

\[ y - 4 = -\frac{3}{4}(x-3) \implies 4y-16 = -3x+9 \implies 3x+4y = 25 \]
Verify the answer using the formula: the tangent to \(x^2+y^2=r^2\) at \((x_1,y_1)\) is \(x_1 x + y_1 y = r^2\).
\[ 3x + 4y = 25 quad \checkmark \]
Determine whether \( (5, -2) \) lies on, inside or outside the circle \( x^2+y^2=30 \).
\[ 5^2 + (-2)^2 = 25 + 4 = 29 < 30 \]

Since \( 29 < 30 \), the point lies inside the circle.

Find the two values of \(y\) such that \( (2, y) \) lies on the circle \(x^2+y^2=29\).
\[ 4 + y^2 = 29 \implies y^2 = 25 \implies y = \pm 5 \]

 Key Takeaways

  • Circle with centre \(O\) and radius \(r\): \( x^2 + y^2 = r^2 \). The radius is \( \sqrt{r^2} \), not \(r^2\).
  • Check whether a point \((a,b)\) lies on the circle by substituting: if \(a^2+b^2=r^2\) → on; less → inside; greater → outside.
  • Tangent at \((x_1, y_1)\) is perpendicular to the radius through that point. Perpendicular gradients multiply to \(-1\).
  • Shortcut: tangent to \(x^2+y^2=r^2\) at \((x_1, y_1)\) has equation \(x_1 x + y_1 y = r^2\).
  • Pythagoras and the circle equation are intimately linked — every point on the circle satisfies \(x^2+y^2=r^2\) precisely because its distance from the origin is \(r\).