Get detailed, step-by-step solutions for NCERT Class 11 Maths Chapter 10 Exercise 10.1 . Learn to find the equation of a circle given its center $(h, k)$ and radius $r$ (Q.1-5, 14). Master converting the general equation to find the center and radius (Q.6-9). Solve problems for circles passing through given points and whose center lies on a specific line (Q.10, 11, 12, 13). Determine if a point lies inside, outside, or on the circle (Q.15).
This exercise involves finding the equation of a circle given its center and radius, and conversely, finding the center and radius from the circle’s equation.
The standard equation of a circle with center $(h, k)$ and radius $r$ is:
$$(x – h)^2 + (y – k)^2 = r^2$$
The general equation of a circle is:
$$x^2 + y^2 + 2gx + 2fy + c = 0$$
where the center is $(-g, -f)$ and the radius is $r = \sqrt{g^2 + f^2 – c}$.
Finding the Equation of the Circle (Exercises 1–5)
We use the standard equation $(x – h)^2 + (y – k)^2 = r^2$.
1. Center $(0, 2)$ and radius $2$.
$$(x – 0)^2 + (y – 2)^2 = 2^2$$
$$\mathbf{x^2 + (y – 2)^2 = 4}$$
2. Center $(-2, 3)$ and radius $4$.
$$(x – (-2))^2 + (y – 3)^2 = 4^2$$
$$\mathbf{(x + 2)^2 + (y – 3)^2 = 16}$$
3. Center $\left(\frac{1}{4}, \frac{1}{2}\right)$ and radius $\frac{1}{12}$.
$$\left(x – \frac{1}{4}\right)^2 + \left(y – \frac{1}{2}\right)^2 = \left(\frac{1}{12}\right)^2$$
$$\mathbf{\left(x – \frac{1}{4}\right)^2 + \left(y – \frac{1}{2}\right)^2 = \frac{1}{144}}$$
4. Center $(1, 1)$ and radius $\sqrt{2}$.
$$(x – 1)^2 + (y – 1)^2 = (\sqrt{2})^2$$
$$\mathbf{(x – 1)^2 + (y – 1)^2 = 2}$$
5. Center $(-a, -b)$ and radius $\sqrt{a^2 – b^2}$.
$$(x – (-a))^2 + (y – (-b))^2 = \left(\sqrt{a^2 – b^2}\right)^2$$
$$(x + a)^2 + (y + b)^2 = a^2 – b^2$$
Expand:
$$(x^2 + 2ax + a^2) + (y^2 + 2by + b^2) = a^2 – b^2$$
$$x^2 + y^2 + 2ax + 2by + a^2 + b^2 – a^2 + b^2 = 0$$
$$\mathbf{x^2 + y^2 + 2ax + 2by + 2b^2 = 0}$$
Finding the Center and Radius (Exercises 6–9)
6. $(x + 5)^2 + (y – 3)^2 = 36$
The equation is in the standard form $(x – h)^2 + (y – k)^2 = r^2$.
$$(x – (-5))^2 + (y – 3)^2 = 6^2$$
- Center $(h, k)$: $\mathbf{(-5, 3)}$
- Radius $r$: $\mathbf{6}$
7. $x^2 + y^2 – 4x – 8y – 45 = 0$
The equation is in the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.
$2g = -4 \implies g = -2$.
$2f = -8 \implies f = -4$.
$c = -45$.
- Center $(-g, -f)$: $(-(-2), -(-4)) = \mathbf{(2, 4)}$
- Radius $r = \sqrt{g^2 + f^2 – c}$:$$r = \sqrt{(-2)^2 + (-4)^2 – (-45)}$$$$r = \sqrt{4 + 16 + 45} = \sqrt{65}$$
- Radius $r$: $\mathbf{\sqrt{65}}$
8. $x^2 + y^2 – 8x + 10y – 12 = 0$
$2g = -8 \implies g = -4$.
$2f = 10 \implies f = 5$.
$c = -12$.
- Center $(-g, -f)$: $(-(-4), -5) = \mathbf{(4, -5)}$
- Radius $r = \sqrt{g^2 + f^2 – c}$:$$r = \sqrt{(-4)^2 + 5^2 – (-12)}$$$$r = \sqrt{16 + 25 + 12} = \sqrt{53}$$
- Radius $r$: $\mathbf{\sqrt{53}}$
9. $2x^2 + 2y^2 – x = 0$
First, divide by 2 to get the coefficient of $x^2$ and $y^2$ equal to 1:
$$x^2 + y^2 – \frac{1}{2}x = 0$$
$2g = -\frac{1}{2} \implies g = -\frac{1}{4}$.
$2f = 0 \implies f = 0$.
$c = 0$.
- Center $(-g, -f)$: $\left(-(-\frac{1}{4}), -0\right) = \mathbf{\left(\frac{1}{4}, 0\right)}$
- Radius $r = \sqrt{g^2 + f^2 – c}$:$$r = \sqrt{\left(-\frac{1}{4}\right)^2 + 0^2 – 0}$$$$r = \sqrt{\frac{1}{16}} = \frac{1}{4}$$
- Radius $r$: $\mathbf{\frac{1}{4}}$
Advanced Problems (Exercises 10–14)
10. Circle passing through $(4, 1)$ and $(6, 5)$, center on $4x + y = 16$.
Let the center be $(h, k)$. Since the center lies on $4x + y = 16$, we have:
$$\mathbf{4h + k = 16} \quad \text{(i)}$$
The distance from the center $(h, k)$ to the two points must be equal (radius $r$):
$$(h – 4)^2 + (k – 1)^2 = (h – 6)^2 + (k – 5)^2$$
$$h^2 – 8h + 16 + k^2 – 2k + 1 = h^2 – 12h + 36 + k^2 – 10k + 25$$
Cancel $h^2$ and $k^2$:
$$-8h – 2k + 17 = -12h – 10k + 61$$
$$4h + 8k = 44$$
$$\mathbf{h + 2k = 11} \quad \text{(ii)}$$
Solve (i) and (ii):
From (i), $k = 16 – 4h$. Substitute into (ii):
$$h + 2(16 – 4h) = 11$$
$$h + 32 – 8h = 11$$
$$-7h = -21 \implies \mathbf{h = 3}$$
Substitute $h=3$ into (i):
$$4(3) + k = 16 \implies 12 + k = 16 \implies \mathbf{k = 4}$$
Center is $(3, 4)$.
Now find the radius $r$: use $(4, 1)$.
$$r^2 = (4 – 3)^2 + (1 – 4)^2 = 1^2 + (-3)^2 = 1 + 9 = 10$$
The equation is $(x – 3)^2 + (y – 4)^2 = 10$.
$$\mathbf{x^2 + y^2 – 6x – 8y + 15 = 0}$$
11. Circle passing through $(2, 3)$ and $(-1, 1)$, center on $x – 3y – 11 = 0$.
Let the center be $(h, k)$. Since the center lies on $x – 3y – 11 = 0$:
$$\mathbf{h – 3k = 11} \quad \text{(i)}$$
Equate the square of the radii to the points $(2, 3)$ and $(-1, 1)$:
$$(h – 2)^2 + (k – 3)^2 = (h – (-1))^2 + (k – 1)^2$$
$$h^2 – 4h + 4 + k^2 – 6k + 9 = h^2 + 2h + 1 + k^2 – 2k + 1$$
$$-4h – 6k + 13 = 2h – 2k + 2$$
$$11 = 6h + 4k$$
$$\mathbf{6h + 4k = 11} \quad \text{(ii)}$$
Solve (i) and (ii):
From (i), $h = 11 + 3k$. Substitute into (ii):
$$6(11 + 3k) + 4k = 11$$
$$66 + 18k + 4k = 11$$
$$22k = 11 – 66 = -55$$
$$k = -\frac{55}{22} = -\frac{5}{2}$$
Substitute $k = -\frac{5}{2}$ into (i):
$$h = 11 + 3\left(-\frac{5}{2}\right) = 11 – \frac{15}{2} = \frac{22 – 15}{2} = \frac{7}{2}$$
Center is $\left(\frac{7}{2}, -\frac{5}{2}\right)$.
Now find the radius $r$: use $(2, 3)$.
$$r^2 = \left(2 – \frac{7}{2}\right)^2 + \left(3 – \left(-\frac{5}{2}\right)\right)^2$$
$$r^2 = \left(\frac{4 – 7}{2}\right)^2 + \left(\frac{6 + 5}{2}\right)^2 = \left(-\frac{3}{2}\right)^2 + \left(\frac{11}{2}\right)^2$$
$$r^2 = \frac{9}{4} + \frac{121}{4} = \frac{130}{4} = \frac{65}{2}$$
The equation is $\left(x – \frac{7}{2}\right)^2 + \left(y + \frac{5}{2}\right)^2 = \frac{65}{2}$.
$$x^2 – 7x + \frac{49}{4} + y^2 + 5y + \frac{25}{4} = \frac{65}{2}$$
$$x^2 + y^2 – 7x + 5y + \frac{74}{4} = \frac{130}{4}$$
$$x^2 + y^2 – 7x + 5y – \frac{56}{4} = 0$$
$$\mathbf{2x^2 + 2y^2 – 14x + 10y – 28 = 0} \quad \text{or} \quad \mathbf{x^2 + y^2 – 7x + 5y – 14 = 0}$$
12. Circle with radius $5$, center on $x$-axis, passes through $(2, 3)$.
Center $(h, k)$ is on the $x$-axis, so $k=0$. Center is $(h, 0)$. Radius $r=5$.
The equation is $(x – h)^2 + (y – 0)^2 = 5^2$.
$$(x – h)^2 + y^2 = 25$$
It passes through $(2, 3)$:
$$(2 – h)^2 + 3^2 = 25$$
$$4 – 4h + h^2 + 9 = 25$$
$$h^2 – 4h + 13 = 25$$
$$h^2 – 4h – 12 = 0$$
Factor: $(h – 6)(h + 2) = 0$.
$$h = 6 \quad \text{or} \quad h = -2$$
Two possible circles:
- Center $(6, 0)$: $(x – 6)^2 + y^2 = 25 \implies \mathbf{x^2 + y^2 – 12x + 11 = 0}$
- Center $(-2, 0)$: $(x + 2)^2 + y^2 = 25 \implies \mathbf{x^2 + y^2 + 4x – 21 = 0}$
13. Circle passing through $(0, 0)$ and making intercepts $a$ and $b$ on the axes.
Since the circle passes through $(0, 0)$, the intercepts $(a, 0)$ and $(0, b)$ are points on the circle.
The points $(0, 0)$, $(a, 0)$, and $(0, b)$ all lie on the circle. Since the angle formed by $(a, 0)$, $(0, 0)$, and $(0, b)$ is $90^\circ$, the segment connecting the intercepts, which is the line segment from $(a, 0)$ to $(0, b)$, must be a diameter of the circle.
The center $(h, k)$ is the midpoint of the diameter:
$$h = \frac{a + 0}{2} = \frac{a}{2}$$
$$k = \frac{0 + b}{2} = \frac{b}{2}$$
Center is $\left(\frac{a}{2}, \frac{b}{2}\right)$.
The radius squared $r^2$ is the distance squared from the center to $(0, 0)$:
$$r^2 = \left(\frac{a}{2} – 0\right)^2 + \left(\frac{b}{2} – 0\right)^2 = \frac{a^2}{4} + \frac{b^2}{4}$$
The equation is:
$$\left(x – \frac{a}{2}\right)^2 + \left(y – \frac{b}{2}\right)^2 = \frac{a^2 + b^2}{4}$$
Expand:
$$x^2 – ax + \frac{a^2}{4} + y^2 – by + \frac{b^2}{4} = \frac{a^2}{4} + \frac{b^2}{4}$$
$$x^2 + y^2 – ax – by = 0$$
$$\mathbf{x^2 + y^2 – ax – by = 0}$$
14. Circle with center $(2, 2)$ and passes through the point $(4, 5)$.
Center $(h, k) = (2, 2)$.
Find $r^2$ using the distance formula between $(2, 2)$ and $(4, 5)$:
$$r^2 = (4 – 2)^2 + (5 – 2)^2$$
$$r^2 = 2^2 + 3^2 = 4 + 9 = 13$$
The equation is $(x – 2)^2 + (y – 2)^2 = 13$.
$$\mathbf{x^2 + y^2 – 4x – 4y – 5 = 0}$$
15. Point Location Relative to a Circle
Does the point $(-2.5, 3.5)$ lie inside, outside or on the circle $x^2 + y^2 = 25$?
The circle is $x^2 + y^2 = 5^2$. Center is $(0, 0)$, and radius is $r = 5$.
A point $(x_1, y_1)$ is:
- Inside if $x_1^2 + y_1^2 < r^2$.
- On if $x_1^2 + y_1^2 = r^2$.
- Outside if $x_1^2 + y_1^2 > r^2$.
Substitute the point $(-2.5, 3.5)$:
$$(-2.5)^2 + (3.5)^2$$
$$6.25 + 12.25 = 18.5$$
Compare $18.5$ with $r^2 = 25$.
Since $18.5 < 25$, the point lies inside the circle.