MHT-CET Maths · Circle

Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms

(x − h)² + (y − k)² = r²; the general form x² + y² + 2gx + 2fy + c = 0 has centre (−g, −f) and radius √(g² + f² − c); the diameter form (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0; and the parametric form x = h + r cos θ, y = k + r sin θ.

All Circle notes MHT-CET Maths strategy

Why this matters

12 PYQs at 33% HARD — the chapter's opening page and its largest. The diameter form is the workhorse: the diagonal of a rectangle (twice), the centres of two circles as the diameter's ends, and the 2023/2024 stem whose endpoints are the roots of two quadratics. The parametric form and the centre-from-two-diameters stems are one line each; the two HARD algebraic ones — four points (m, 1/m) on a circle and a count of integral k for a bounded radius — are the general equation read as a polynomial.

Concept 1 of 4

Centre-Radius and General Forms: Read (−g, −f) and √(g² + f² − c)

Intuition

Expand

(x - h)^2 + (y - k)^2 = r^2

and you get the general form with

g = -h

f = -k

c = h^2 + k^2 - r^2

. Going back is completing the square. The centre is the intersection of any two diameters, and a circle through two points with its centre on a known line is fixed by equating two distances.

Definition

Diameters along $3x - 4y = 7$ and $2x - 3y = 5$ meet at $(1, -1)$ ; area $49\pi$ gives $r = 7$ : $x^2 + y^2 - 2x + 2y - 47 = 0$ .
Centre $(0, a)$ on the $y$ -axis, through $(4, 0)$ and $(0, 2)$ : $16 + a^2 = (a - 2)^2 \Rightarrow a = 3$ , $r = 5$ ; $r^2 - r + 1 = 21$ .
Two diameters given as a pair $x^2 - y^2 - 2x + 4y - 3 = (x + y - 3)(x - y + 1)$ : centre $(1, 2)$ ; through $(1, 1)$ means $r = 1$ .
$x^2 + y^2 + kx + (1 - k)y + 5 = 0$ , radius at most $5$ : $\dfrac{k^2 + (1-k)^2}{4} - 5 \le 25 \Rightarrow 2k^2 - 2k - 119 \le 0 \Rightarrow -7.23 \le k \le 8.23$ , sixteen integers — the official key. Strictly the radius must also be REAL ( $g^2 + f^2 - c > 0$ ), which drops $k = -2, \dots, 3$ and leaves ten; that count is not offered.
The coefficients of $x^2$ and $y^2$ must be equal (scale to $1$ ) and there must be no $xy$ term; otherwise it is not a circle.

Two forms

(x - h)^2 + (y - k)^2 = r^2 \qquad x^2 + y^2 + 2gx + 2fy + c = 0:\ C(-g, -f),\ r = \sqrt{g^2 + f^2 - c}

Worked example

Find the centre and radius of

2x^2 + 2y^2 - 8x + 12y - 6 = 0

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1CircleMODERATE

If the lines

3x-4y-7=0

and

2x-3y-5=0

pass through diameters of a circle of area

49\pi

square units, then the equation of the circle is

[Q145 · Shift 1 · 2022]

Reading the centre as (g, f)

x^2 + y^2 - 4x + 6y - 3 = 0

has centre

(2, -3)

: the signs FLIP. Both sign-error centres are always on the option list.

Drill 3 more on centre-radius and general forms: read (−g, −f) and √(g² + f² − c)

Concept 2 of 4

Diameter Form: (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0

Intuition

The angle in a semicircle is a right angle, so a point

P

is on the circle with diameter

AB

iff

PA \perp PB

; the dot product of the two direction vectors is the diameter form. It never needs the centre or radius.

Definition

Rectangle $x = -2, x = 6, y = -2, y = 5$ : a diagonal joins $(-2, -2)$ , $(6, 5)$ : $(x + 2)(x - 6) + (y + 2)(y - 5) = x^2 + y^2 - 4x - 3y - 22 = 0$ .
Centres $(-3, 7)$ and $(2, -5)$ as ends: $(x + 3)(x - 2) + (y - 7)(y + 5) = x^2 + y^2 + x - 2y - 41 = 0$ .
Abscissae the roots of $x^2 + 2ax - b^2 = 0$ , ordinates of $y^2 + 2py - q^2 = 0$ : expanding the diameter form gives $x^2 - (x_1 + x_2)x + x_1x_2 + y^2 - (y_1 + y_2)y + y_1y_2 = 0$ , and Vieta fills in: $x^2 + y^2 + 2ax + 2py - b^2 - q^2 = 0$ .
The expansion needs only the SUM and PRODUCT of the coordinates — which is why the roots-of-quadratics stem never asks you to solve the quadratics.

Diameter form

(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \iff x^2 + y^2 - (x_1 + x_2)x - (y_1 + y_2)y + x_1x_2 + y_1y_2 = 0

Worked example

Find the circle with

(1, 2)

and

(5, -4)

as the ends of a diameter.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2CircleHARD

The abscissae of the two points A and B are the roots of the equation

x^2+2ax-b^2=0

and their ordinates are roots of the equation

y^2+2py-q^2=0

. Then the equation of the circle with AB as diameter is given by

[Q109 · 10th May Shift 1 · 2023]

Sign of the constant with negative products

x_1x_2 + y_1y_2 = -b^2 - q^2

. The option with

-b^2 + q^2

is the planted slip; the paper prints

-(b^2 + q^2)

Drill 4 more on diameter form: (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0

Concept 3 of 4

Parametric Form: x = h + r cos θ, y = k + r sin θ

Intuition

Every point of a circle is the centre plus a radius-length step at angle

\theta

. Complete the square to find

(h, k)

and

r

, then write the two coordinates.

Definition

$x^2 + y^2 + 2x - 4y - 4 = 0 \Rightarrow (x + 1)^2 + (y - 2)^2 = 9$ : $x = -1 + 3\cos\theta$ , $y = 2 + 3\sin\theta$ .
$x^2 + y^2 - ax - by = 0$ : centre $\left(\dfrac{a}{2}, \dfrac{b}{2}\right)$ , $r = \dfrac{\sqrt{a^2 + b^2}}{2}$ .
$\cos$ goes with $x$ and $\sin$ with $y$ ; the swapped version is a circle too, but not the standard parametrisation the options test.
$x = 5\cos\theta$ , $y = 5\sin\theta$ is $x^2 + y^2 = 25$ ; the tangent at parameter $\theta$ is $x\cos\theta + y\sin\theta = 5$ .

Parametric circle

x = h + r\cos\theta,\quad y = k + r\sin\theta \qquad (0 \le \theta < 2\pi)

Worked example

Write parametric equations for

x^2 + y^2 - 6x + 8y = 0

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3CircleEASY

The parametric equation of the circle

x^2+y^2+2x-4y-4=0

are

[Q130 · 10th May Shift 2 · 2024]

Halving the radius with the centre

For

x^2 + y^2 - ax - by = 0

the centre halves

a, b

and the radius is

\dfrac{\sqrt{a^2 + b^2}}{2}

, not

\dfrac{\sqrt{a^2 + b^2}}{4}

. Options (B) and (D) carry the quartered radius.

Drill 1 more on parametric form: x = h + r cos θ, y = k + r sin θ

Concept 4 of 4

Points of a Family on a Circle: Substitute, Get a Polynomial, Use Vieta

Intuition

If points of the form

\left(m, \dfrac1m\right)

lie on a circle, substituting into the general equation gives one polynomial in

m

whose roots are the parameters of the points. Multiply out the denominators and read off the product of the roots.

Definition

$\left(m, \dfrac1m\right)$ on $x^2 + y^2 + 2gx + 2fy + c = 0$ : $m^2 + \dfrac{1}{m^2} + 2gm + \dfrac{2f}{m} + c = 0 \Rightarrow m^4 + 2gm^3 + cm^2 + 2fm + 1 = 0$ . Product of the four roots $= \dfrac{1}{1} = 1$ .
For a quartic $m^4 + \alpha m^3 + \beta m^2 + \gamma m + \delta = 0$ , the product of the roots is $+\delta$ (even degree) and the sum is $-\alpha$ .
The same substitution answers 'how many points of the form … lie on the circle' — the degree of the polynomial bounds the count.

Vieta on the substituted polynomial

m^4 + 2gm^3 + cm^2 + 2fm + 1 = 0 \ \Rightarrow\ m_1 m_2 m_3 m_4 = 1

Worked example

Four points

\left(t, \dfrac2t\right)

lie on a circle. Find the product of the four values of

t

Practice this conceptself-check

From the bank · past-year question

Example 4CircleHARD

\left(m_i, \dfrac{1}{m_i}\right)

m_i > 0

i = 1, 2, 3, 4

are four distinct points on a circle, then the product

m_1 m_2 m_3 m_4

is equal to

[Q130 · 2nd May Shift 1 · 2023]

Product of roots with the wrong sign

For an even-degree monic polynomial the product of the roots is

+

(constant term);

-1

is option (A) and comes from applying the cubic's sign rule.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Centre-Radius and General Forms: Read (−g, −f) and √(g² + f² − c)
Two forms
$(x - h)^2 + (y - k)^2 = r^2 \qquad x^2 + y^2 + 2gx + 2fy + c = 0:\ C(-g, -f),\ r = \sqrt{g^2 + f^2 - c}$
Diameter Form: (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0
Diameter form
$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \iff x^2 + y^2 - (x_1 + x_2)x - (y_1 + y_2)y + x_1x_2 + y_1y_2 = 0$
Parametric Form: x = h + r cos θ, y = k + r sin θ
Parametric circle
$x = h + r\cos\theta,\quad y = k + r\sin\theta \qquad (0 \le \theta < 2\pi)$
Points of a Family on a Circle: Substitute, Get a Polynomial, Use Vieta
Vieta on the substituted polynomial
$m^4 + 2gm^3 + cm^2 + 2fm + 1 = 0 \ \Rightarrow\ m_1 m_2 m_3 m_4 = 1$

Watch out for (4)

Reading the centre as (g, f)
→ Centre-Radius and General Forms: Read (−g, −f) and √(g² + f² − c)
Sign of the constant with negative products
→ Diameter Form: (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0
Halving the radius with the centre
→ Parametric Form: x = h + r cos θ, y = k + r sin θ
Product of roots with the wrong sign
→ Points of a Family on a Circle: Substitute, Get a Polynomial, Use Vieta

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