NCERT Class 11 Mathematics — Conic Sections, Exercise 10.1. All 15 questions solved.

A circle is the set of points at a fixed distance $r$ (the radius) from a fixed point $(h, k)$ (the centre). Applying the distance formula gives the standard form:

$$(x – h)^2 + (y – k)^2 = r^2$$

Expanding produces the general form:

$$x^2 + y^2 + 2gx + 2fy + c = 0, \qquad \text{centre } (-g, -f), \qquad r = \sqrt{g^2 + f^2 – c}$$

Moving between the two is done by completing the square, and that single technique answers questions 6 to 9.

Key insight. When a circle must pass through given points and have its centre on a given line, write the centre as $(h, k)$ and collect equations: each point supplies one (by equating distances to the centre), and the line supplies one more. Two points plus a line is exactly enough to pin down $h$, $k$ and then $r$ — questions 10 and 11 are that recipe.

In questions 1 to 5, find the equation of the circle with the given centre and radius.

Question 1

Centre $(0, 2)$ and radius $2$.

Solution.

$$(x – 0)^2 + (y – 2)^2 = 2^2 \quad\Longrightarrow\quad x^2 + y^2 – 4y + 4 = 4$$

$$x^2 + y^2 – 4y = 0$$

Question 2

Centre $(-2, 3)$ and radius $4$.

Solution.

$$(x + 2)^2 + (y – 3)^2 = 16$$

$$x^2 + 4x + 4 + y^2 – 6y + 9 = 16 \quad\Longrightarrow\quad x^2 + y^2 + 4x – 6y – 3 = 0$$

Question 3

Centre $\left(\dfrac12, \dfrac14\right)$ and radius $\dfrac{1}{12}$.

Solution.

$$\left(x – \frac12\right)^2 + \left(y – \frac14\right)^2 = \frac{1}{144}$$

$$x^2 – x + \frac14 + y^2 – \frac y2 + \frac{1}{16} = \frac{1}{144}$$

Multiply throughout by $144$ to clear all fractions:

$$144x^2 – 144x + 36 + 144y^2 – 72y + 9 = 1$$

$$144x^2 + 144y^2 – 144x – 72y + 44 = 0$$

Dividing by $4$:

$$36x^2 + 36y^2 – 36x – 18y + 11 = 0$$

Question 4

Centre $(1, 1)$ and radius $\sqrt2$.

Solution.

$$(x-1)^2 + (y-1)^2 = 2$$

$$x^2 – 2x + 1 + y^2 – 2y + 1 = 2 \quad\Longrightarrow\quad x^2 + y^2 – 2x – 2y = 0$$

The constant term vanishing means the circle passes through the origin — which checks out, since the distance from $(1,1)$ to $(0,0)$ is $\sqrt2$.

Question 5

Centre $(-a, -b)$ and radius $\sqrt{a^2 – b^2}$.

Solution.

$$(x + a)^2 + (y + b)^2 = a^2 – b^2$$

$$x^2 + 2ax + a^2 + y^2 + 2by + b^2 = a^2 – b^2$$

The $a^2$ cancels, and $b^2$ moves across to become $2b^2$:

$$x^2 + y^2 + 2ax + 2by + 2b^2 = 0$$

In questions 6 to 9, find the centre and radius of the circles.

Question 6

$(x + 5)^2 + (y – 3)^2 = 36$

Solution. Already in standard form. Comparing with $(x-h)^2 + (y-k)^2 = r^2$:

$$h = -5, \qquad k = 3, \qquad r = \sqrt{36} = 6$$

Note the sign flip: $(x + 5)^2$ means $h = -5$, not $+5$.

Question 7

$x^2 + y^2 – 4x – 8y – 45 = 0$

Solution. Complete the square in $x$ and in $y$ separately.

$$\left(x^2 – 4x\right) + \left(y^2 – 8y\right) = 45$$

Half of $-4$ is $-2$, and half of $-8$ is $-4$:

$$\left(x – 2\right)^2 – 4 + \left(y – 4\right)^2 – 16 = 45$$

$$(x-2)^2 + (y-4)^2 = 65$$

Question 8

$x^2 + y^2 – 8x + 10y – 12 = 0$

Solution.

$$\left(x^2 – 8x\right) + \left(y^2 + 10y\right) = 12$$

$$(x – 4)^2 – 16 + (y + 5)^2 – 25 = 12$$

$$(x-4)^2 + (y+5)^2 = 53$$

Question 9

$2x^2 + 2y^2 – x = 0$

Solution. The coefficients of $x^2$ and $y^2$ must be $1$ before completing the square, so divide throughout by $2$ first:

$$x^2 + y^2 – \frac x2 = 0$$

$$\left(x^2 – \frac x2\right) + y^2 = 0$$

Half of $-\tfrac12$ is $-\tfrac14$:

$$\left(x – \frac14\right)^2 – \frac{1}{16} + y^2 = 0 \quad\Longrightarrow\quad \left(x – \frac14\right)^2 + y^2 = \frac{1}{16}$$

Question 10

Find the equation of the circle passing through the points $(4, 1)$ and $(6, 5)$ and whose centre is on the line $4x + y = 16$.

Solution. Let the centre be $(h, k)$. Three facts are available.

The centre is equidistant from the two points:

$$(4-h)^2 + (1-k)^2 = (6-h)^2 + (5-k)^2$$

$$16 – 8h + 1 – 2k = 36 – 12h + 25 – 10k$$

$$4h + 8k = 44 \quad\Longrightarrow\quad h + 2k = 11 \qquad (1)$$

The centre lies on the given line:

$$4h + k = 16 \qquad (2)$$

From (1), $h = 11 – 2k$; substituting into (2):

$$44 – 8k + k = 16 \quad\Longrightarrow\quad -7k = -28 \quad\Longrightarrow\quad k = 4, \qquad h = 3$$

The radius:

$$r^2 = (4-3)^2 + (1-4)^2 = 1 + 9 = 10$$

$$(x-3)^2 + (y-4)^2 = 10 \quad\Longrightarrow\quad x^2 + y^2 – 6x – 8y + 15 = 0$$

Question 11

Find the equation of the circle passing through the points $(2, 3)$ and $(-1, 1)$ and whose centre is on the line $x – 3y – 11 = 0$.

Solution. With centre $(h, k)$, equidistance gives

$$(2-h)^2 + (3-k)^2 = (-1-h)^2 + (1-k)^2$$

$$4 – 4h + 9 – 6k = 1 + 2h + 1 – 2k$$

$$6h + 4k = 11 \qquad (1)$$

The line gives $h = 3k + 11$. Substituting into (1):

$$6(3k+11) + 4k = 11 \quad\Longrightarrow\quad 22k = -55 \quad\Longrightarrow\quad k = -\frac52, \qquad h = \frac72$$

$$r^2 = \left(2 – \frac72\right)^2 + \left(3 + \frac52\right)^2 = \frac94 + \frac{121}{4} = \frac{130}{4} = \frac{65}{2}$$

$$\left(x – \frac72\right)^2 + \left(y + \frac52\right)^2 = \frac{65}{2}$$

Expanding:

$$x^2 – 7x + \frac{49}{4} + y^2 + 5y + \frac{25}{4} = \frac{130}{4}$$

$$x^2 + y^2 – 7x + 5y – 14 = 0$$

Question 12

Find the equation of the circle with radius $5$ whose centre lies on the $x$-axis and passes through the point $(2, 3)$.

Solution. A centre on the $x$-axis has the form $(a, 0)$. The distance to $(2,3)$ must be $5$:

$$(2 – a)^2 + 9 = 25 \quad\Longrightarrow\quad (2-a)^2 = 16 \quad\Longrightarrow\quad 2 – a = \pm4$$

$$a = -2 \qquad\text{or}\qquad a = 6$$

Both give valid circles — the point $(2,3)$ sits at distance $5$ from each centre.

$a = -2$: $(x+2)^2 + y^2 = 25 \Rightarrow x^2 + y^2 + 4x – 21 = 0$

$a = 6$: $(x-6)^2 + y^2 = 25 \Rightarrow x^2 + y^2 – 12x + 11 = 0$

Question 13

Find the equation of the circle passing through $(0, 0)$ and making intercepts $a$ and $b$ on the coordinate axes.

Solution. Making an intercept $a$ on the $x$-axis and $b$ on the $y$-axis, while also passing through the origin, means the circle goes through three points:

$$(0,0), \qquad (a, 0), \qquad (0, b)$$

Take the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.

Through $(0,0)$: substituting gives $c = 0$ immediately — which is why starting from the general form pays here.

Through $(a, 0)$: $a^2 + 2ga = 0$, so $a(a + 2g) = 0$ and (for $a \ne 0$) $g = -\tfrac a2$.

Through $(0, b)$: similarly $f = -\tfrac b2$.

$$x^2 + y^2 – ax – by = 0$$

Question 14

Find the equation of a circle with centre $(2, 2)$ and passing through the point $(4, 5)$.

Solution. The radius is the distance from the centre to the point:

$$r^2 = (4-2)^2 + (5-2)^2 = 4 + 9 = 13$$

$$(x-2)^2 + (y-2)^2 = 13$$

Expanding:

$$x^2 – 4x + 4 + y^2 – 4y + 4 = 13 \quad\Longrightarrow\quad x^2 + y^2 – 4x – 4y = 5$$

Question 15

Does the point $(-2.5, 3.5)$ lie inside, outside or on the circle $x^2 + y^2 = 25$?

Solution. The circle is centred at the origin with radius $5$, so compare the point’s distance from the origin with $5$. Working with squares avoids the radical:

$$(-2.5)^2 + (3.5)^2 = 6.25 + 12.25 = 18.5$$

Since $18.5 < 25$, the point is closer to the centre than the radius:

$$\text{distance} = \sqrt{18.5} \approx 4.30 < 5$$

Common mistakes

• Question 6, reading the centre with the wrong signs. $(x+5)^2$ corresponds to $h = -5$, because the standard form is $(x – h)^2$. Rewriting as $\left(x – (-5)\right)^2$ makes it unambiguous.
• Question 9, completing the square before dividing. The general form requires the coefficients of $x^2$ and $y^2$ to be $1$. Dividing by $2$ first is essential; completing the square on $2x^2 – x$ directly gives a wrong radius.
• Completing the square with the wrong constant. For $x^2 – 4x$, add and subtract $\left(\tfrac{-4}{2}\right)^2 = 4$, not $16$. Halving then squaring is the order.
• Question 12, taking only one root. $(2-a)^2 = 16$ has two solutions, giving two circles that both satisfy every condition stated.
• Question 13, assuming the intercepts are $\tfrac a2$ and $\tfrac b2$. The circle meets the axes at $(a,0)$ and $(0,b)$ — those are the intercepts. The halves appear in $g$ and $f$, not in the points.
• Question 15, comparing $18.5$ with $5$. The quantity $18.5$ is the squared distance, so it must be compared with $r^2 = 25$, or square-rooted first. Comparing it with $5$ would wrongly place the point outside.
• Forgetting that $g^2 + f^2 – c$ must be positive. If it were negative the equation would describe no real points at all, and if zero, a single point.

Practise next

• Exercise 10.2 — parabolas, where the focus-directrix definition replaces the centre-radius one.
• Exercise 9.1 — worth revising if the distance and midpoint calculations in questions 10 to 12 felt slow.