Circles centred at the origin and their tangents

WJEC Higher has a focused circle topic: the circle x² + y² = r².

The standard form

A circle with centre (0, 0) and radius r has equation: x² + y² = r².

So x² + y² = 25 has centre (0, 0) and radius 5.

If the equation appears as x² + y² = c with c < 0, it has no real solutions (no graph). If c = 0 it is just the single point (0, 0).

Finding the radius from a point on the circle

If (3, 4) lies on x² + y² = r², then r² = 9 + 16 = 25, so r = 5.

Tangent to a circle at a point P

A tangent at P touches the circle and is PERPENDICULAR to the radius OP.

Algorithm:

1. Compute the gradient of OP: m_OP = y₁ / x₁.
2. The tangent gradient is the negative reciprocal: m_T = − x₁ / y₁.
3. Use point-gradient form through P: y − y₁ = m_T (x − x₁).

Special cases

• Tangent at (r, 0) or (−r, 0): gradient is undefined (vertical). Equation is x = r or x = −r.
• Tangent at (0, r) or (0, −r): gradient zero (horizontal). Equation is y = r or y = −r.

Verifying a tangent

A tangent line meets the circle at exactly ONE point. To verify, substitute the line equation into the circle equation; the resulting quadratic should have a discriminant of 0.

WJEC exam tip

Higher tangent questions often present P as an obvious lattice point on the circle (e.g. (3, 4) on radius-5 circle, or (5, 12) on radius-13). Recognising 3-4-5 and 5-12-13 triples saves time and avoids decimal slips.