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

22 PYQs at 23% 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)

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

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2022 · Shift 1 · Q145Moderate

Example 1 · Circle · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms

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

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

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.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 1 · Q109Hard

Example 2 · Circle · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms

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

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

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

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 10th May Shift 2 · Q130Easy

Example 3 · Circle · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms

The parametric equation of the circle

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

are

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

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

Practise this conceptself-check

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 1 · Q130Hard

Example 4 · Circle · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms

\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

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

Test yourself on Circle

15 past MHT-CET questions from this chapter, timed at 27 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Circle chapter test Drill all 22 questions

Related notes

Pair of Straight Lines — factorising the two diameters →