MHT-CET Maths · Circle

Distance From a Point to a Circle — Greatest, Least, a Line Cutting the Circle and the Segment Area

The least and greatest distances from an external point to a circle are d − r and d + r (d = distance to the centre); a line cuts the circle when its distance from the centre is less than r; and the area a chord cuts off is a sector minus a triangle.

All Circle notes MHT-CET Maths strategy

Why this matters

9 PYQs, none HARD. Greatest and least distance from a point to a circle (twice, once continued to the far end of the diameter), the maximum distance from a point of the circle to a line, a count of integer m for which a line cuts the circle, and the median of an equilateral triangle inscribed in a circle. The first three are the Complex Numbers 'greatest and least modulus' move in coordinate dress.

Concept 1 of 3: Greatest and Least Distance: d ± r From a Point, and From the Circle to a Line

Join the point to the centre; the line through both meets the circle at the nearest and farthest points, at

d - r

and

d + r

. Similarly the farthest point of the circle from a line is the centre's distance to the line plus

r

Definition

$P(2, -7)$ , circle centre $(7, 5)$ , $r = 15$ : $d = 13$ ; least $15 - 13 = 2$ (the point is inside!), greatest $28$ . Inside or outside, the two values are $|d - r|$ and $d + r$ .
$A(10, 7)$ , centre $(2, 1)$ , $r = 5$ : $d = 10$ ; $AM = 5$ , and $AM' = AM + 2r = 15$ where $MM'$ is the diameter through $M$ .
Circle centre $(-1, -1)$ , $r = \sqrt5$ ; line $2x + y + 13 = 0$ at distance $\dfrac{10}{\sqrt5} = 2\sqrt5$ from the centre: maximum distance of a point of the circle from the line $= 3\sqrt5$ , minimum $\sqrt5$ .
Same move as the greatest and least $|z|$ on $|z - a| \le r$ .

Extreme distances

\min = |d - r|,\quad \max = d + r \qquad (d = \text{distance from the point, or the line, to the centre})

Worked example

Find the greatest and least distances of

(7, 9)

from

x^2 + y^2 - 2x - 2y - 7 = 0

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 21 April Shift I · Q147Moderate

Example 1 · Circle · Distance From a Point to a Circle — Greatest, Least, a Line Cutting the Circle and the Segment Area

The minimum distance and maximum distance of the point

P(2, - 7)

from the circle

x^{2}+y^{2}- 14x- 10y- 151 = 0

are respectively

\_\_\_\_

units

Adding r twice for the far end of the diameter

AM' = AM + 2r

because

MM'

is a whole diameter —

5 + 10 = 15

AM + r = 10

is option (A).

Drill 2 more on greatest and least distance: d ± r from a point, and from the circle to a line

Concept 2 of 3: When a Line Cuts the Circle: Distance From the Centre < r

Compare the centre's distance from the line with the radius: less means two intersection points, equal means a tangent, more means no intersection. For a line with a parameter, the inequality gives the range of the parameter.

Definition

$x - 2y = m$ with $x^2 + y^2 = 2x + 4y$ (centre $(1, 2)$ , $r = \sqrt5$ ): $\dfrac{|1 - 4 - m|}{\sqrt5} < \sqrt5 \Rightarrow |m + 3| < 5 \Rightarrow -8 < m < 2$ : nine integers.
Strict inequality for two DISTINCT points; the endpoints $m = -8, 2$ are tangents and are excluded.
The chord length when the line cuts is $2\sqrt{r^2 - p^2}$ , $p$ the centre's distance.

Line and circle

p = \frac{|ah + bk + c|}{\sqrt{a^2 + b^2}}:\quad p < r \text{ cuts},\ p = r \text{ touches},\ p > r \text{ misses};\qquad \text{chord} = 2\sqrt{r^2 - p^2}

Worked example

For how many integers

k

does

y = x + k

cut

x^2 + y^2 = 8

at two distinct points?

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 1 · Q108Moderate

Example 2 · Circle · Distance From a Point to a Circle — Greatest, Least, a Line Cutting the Circle and the Segment Area

If the line

x-2y=m

(

m\in\mathbb{Z}

) intersects the circle

x^2+y^2=2x+4y

at two distinct points, then the number of possible values of

m

are

Counting the tangent cases

'Two distinct points' is strict:

m = -8

and

m = 2

are tangents. Including them gives

11

, option (D); excluding both gives

9

Concept 3 of 3: Area Cut Off by a Chord, and the Circumcircle of an Equilateral Triangle

A chord splits the disc into two segments; the minor one is a sector minus an isosceles triangle, or the integral

2\int \sqrt{a^2 - x^2}\,dx

beyond the chord. For an equilateral triangle the circumradius is two-thirds of the median, so the median is

\dfrac{3R}{2}

Definition

$x^2 + y^2 = a^2$ cut by $x = \dfrac{a}{\sqrt2}$ : the chord subtends $90^\circ$ at the centre; segment $= \dfrac{\pi a^2}{4} - \dfrac{a^2}{2} = \dfrac{a^2}{2}\left(\dfrac{\pi}{2} - 1\right)$ . By integration: $2\int_{a/\sqrt2}^{a}\sqrt{a^2 - x^2}\,dx$ gives the same.
Circle centred at the origin through $A(2, 4)$ : $R = 2\sqrt5$ ; an inscribed equilateral triangle has median $= \dfrac32 R = 3\sqrt5$ .
Segment by a chord subtending angle $\theta$ at the centre: $\dfrac{r^2}{2}(\theta - \sin\theta)$ .
For an equilateral triangle the centroid, circumcentre and orthocentre coincide, and the centroid divides each median $2 : 1$ .

Segment and circumradius

\text{segment} = \frac{r^2}{2}(\theta - \sin\theta) \qquad \text{equilateral: median} = \frac{3R}{2}

Worked example

Find the area of the minor segment cut off from

x^2 + y^2 = 16

by the line

y = 2

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 23 April Shift I · Q105Moderate

Example 3 · Circle · Distance From a Point to a Circle — Greatest, Least, a Line Cutting the Circle and the Segment Area

Let the circle with center at origin pass through the vertices of an equilateral triangle ABC . If

A\equiv (2,4)

, then the length of the median through A is

Taking R as the median

The circumradius is two-thirds of the median, so the median is

\dfrac{3}{2}R = 3\sqrt5

;

2\sqrt5

itself is option (A).

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 (3)

Greatest and Least Distance: d ± r From a Point, and From the Circle to a Line
Extreme distances
$\min = |d - r|,\quad \max = d + r \qquad (d = \text{distance from the point, or the line, to the centre})$
When a Line Cuts the Circle: Distance From the Centre < r
Line and circle
$p = \frac{|ah + bk + c|}{\sqrt{a^2 + b^2}}:\quad p < r \text{ cuts},\ p = r \text{ touches},\ p > r \text{ misses};\qquad \text{chord} = 2\sqrt{r^2 - p^2}$
Area Cut Off by a Chord, and the Circumcircle of an Equilateral Triangle
Segment and circumradius
$\text{segment} = \frac{r^2}{2}(\theta - \sin\theta) \qquad \text{equilateral: median} = \frac{3R}{2}$

Watch out for (3)

Adding r twice for the far end of the diameter
→ Greatest and Least Distance: d ± r From a Point, and From the Circle to a Line
Counting the tangent cases
→ When a Line Cuts the Circle: Distance From the Centre < r
Taking R as the median
→ Area Cut Off by a Chord, and the Circumcircle of an Equilateral Triangle

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