Coordinate Geometry of the Circle

This is the new topic in the 2025–2027 syllabus, and new topics get examined. With no questions in older past papers, most students under-practise it, which makes it a differentiator: prepared candidates collect marks here that the cohort drops. The good news: it’s straight-line geometry plus quadratic machinery you already own.

The equation of a circle

Centre $(a, b)$, radius $r$:

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

Expanded general form: $x^2 + y^2 + 2gx + 2fy + c = 0$, centre $(-g, -f)$, radius $\sqrt{g^2 + f^2 - c}$. Moving between forms is completing the square twice:

$x^2 + y^2 - 6x + 4y - 12 = 0$ $(x - 3)^2 - 9 + (y + 2)^2 - 4 - 12 = 0$ $(x - 3)^2 + (y + 2)^2 = 25 \to$ centre $(3, -2)$, radius $5$

Show both completed squares, each is a method mark, and the read-off centre/radius are the accuracy marks. Sign discipline: from $(y + 2)^2$, the centre’s $y$-coordinate is $-2$.

Building equations from given information uses the toolkit: centre $+$ one point on the circle $\to r^2$ from the length formula; endpoints of a diameter $\to$ centre is the midpoint, radius half the length.

Lines and circles: intersect, touch, miss

Substitute the line into the circle equation and collect, a quadratic appears, and its discriminant tells the whole story: $b^2 - 4ac > 0$ two intersection points (chord), $= 0$ tangent, $< 0$ no contact. This is the same routine as line–curve intersection, so the working habits carry over directly: bracketed substitution line, collect to zero, discriminant stated explicitly, conclusion in words.

Find the points where $y = x + 1$ meets $x^2 + y^2 = 25$: $x^2 + (x + 1)^2 = 25 \to 2x^2 + 2x - 24 = 0 \to x^2 + x - 12 = 0 \to (x + 4)(x - 3) = 0$ $x = -4 \to y = -3$; $x = 3 \to y = 4 \to$ $(-4, -3)$ and $(3, 4)$ ($y$ from the linear equation, as pairs)

Tangents and circle properties

The coordinate versions of the geometry facts do the heavy lifting:

• Tangent $\perp$ radius at the point of contact. Tangent at $P$ on a circle with centre $C$: gradient of $CP$, negative reciprocal, then $y - y_1 = m(x - x_1)$ through $P$. Three steps, three marks.
• Centre lies on the perpendicular bisector of any chord, the route to finding centres from chord data.
• Angle in a semicircle is $90°$. “show $AB$ is a diameter” often reduces to showing a right angle or showing the midpoint of $AB$ is the centre.

For “show the line is a tangent”: discriminant $= 0$ (familiar, safe), or perpendicular distance from centre $=$ radius. Either earns full credit; pick the one you’ve drilled.

Worked exam-style question

Find the equation of the tangent to the circle $x^2 + y^2 - 4x - 6y + 8 = 0$ at the point $P(1, 5)$.

Complete the square: $(x - 2)^2 + (y - 3)^2 = 5 \to$ centre $C(2, 3)$ (M, A) Gradient $CP = \frac{5 - 3}{1 - 2} = -2$ (M) Tangent gradient $= \frac{1}{2}$ (M, negative reciprocal, tangent $\perp$ radius stated) Tangent: $y - 5 = \frac{1}{2}(x - 1) \to$ $2y = x + 9$ (A)

Five marks, and every one sits on a written step, the show-your-working economy in miniature.

Common mistakes in this topic

• Centre signs flipped when reading from $(x + g)^2$ forms or the general equation
• Radius left as $r^2$ (the equation gives r-squared; the question asked for $r$)
• Tangent attempted through the centre instead of the contact point
• Discriminant route abandoned half-collected (must be brought to ”= 0” form first)
• Practising only from old papers, this topic isn’t in them; use specimen and recent sessions (past-paper guide)

New topic, no past-paper safety net, exactly the situation where guided practice pays most. Free 1-hour trial class with Teacher Rig: message us on WhatsApp.