CBSE Class 11 Mathematics Conic Sections Notes Set 01

Important Notes and Formulas

Definition

A circle is the locus of a point whose distance from a fixed point (called the centre) is always constant (called the radius).

Equation of a Circle in Various Forms

(a) The circle with centre at the origin and radius r has the equation x² + y² = r².

(b) The circle with centre (h, k) and radius r has the equation (x − h)² + (y − k)² = r².

(c) The general equation of a circle is x² + y² + 2gx + 2fy + c = 0, with centre (−g, −f) and radius \( \sqrt{g^2 + f^2 - c} \). If:
g² + f² − c > 0 ⇒ real circle
g² + f² − c = 0 ⇒ point circle
g² + f² − c < 0 ⇒ imaginary circle, with real centre, that is (−g, −f)
Note: Every second degree equation in x and y, in which the coefficient of x² is equal to the coefficient of y² and the coefficient of xy is zero, always represents a circle.

(d) The equation of the circle with (x₁, y₁) and (x₂, y₂) as the extremities of a diameter is:
(x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0
Note that this is the circle of least radius passing through (x₁, y₁) and (x₂, y₂).

Intercepts Made by a Circle on the Axes

The intercepts made by the circle x² + y² + 2gx + 2fy + c = 0 on the coordinate axes are \( 2\sqrt{g^2 - c} \) (on the x-axis) and \( 2\sqrt{f^2 - c} \) (on the y-axis) respectively. If:
g² − c > 0 ⇒ the circle cuts the x-axis at two distinct points
g² − c = 0 ⇒ the circle touches the x-axis
g² − c < 0 ⇒ the circle lies completely above or below the x-axis

Position of a Point with respect to a Circle

The point (x₁, y₁) lies inside, on or outside the circle S ≡ x² + y² + 2gx + 2fy + c = 0 according as
S₁ ≡ x₁² + y₁² + 2gx₁ + 2fy₁ + c < 0, = 0 or > 0.

Solved Examples

Example 1. In each of the following questions, find the centre and radius of the circles
(i) (x + 5)² + (y − 3)² = 36
(ii) x² + y² − 4x − 8y − 45 = 0
(iii) x² + y² − 8x + 10y − 12 = 0
(iv) 2x² + 2y² − x = 0
Answer:
(i) (x + 5)² + (y − 3)² = 36
or (x − (−5))² + (y − 3)² = 6²
h = −5, k = 3 and r = 6
Therefore, the given circle has centre at (−5, 3) and radius 6.
(ii) The given equation is x² + y² − 4x − 8y − 45 = 0
or (x² − 4x) + (y² − 8y) = 45
Now completing the squares within the parentheses, we get
(x² − 4x + 4) + (y² − 8y + 16) = 4 + 16 + 45
or (x − 2)² + (y − 4)² = 65
Therefore, the given circle has centre at (2, 4) and radius \( \sqrt{65} \).
(iii) The given equation is x² + y² − 8x + 10y − 12 = 0
or (x² − 8x) + (y² + 10y) = 12
or (x² − 8x + 16) + (y² + 10y + 25) = 12 + 16 + 25
or (x − 4)² + (y + 5)² = 53
Therefore, the given circle has centre at (4, −5) and radius \( \sqrt{53} \).
(iv) The given equation is 2x² + 2y² − x = 0
or \( x^2 + y^2 - \frac{x}{2} = 0 \)
or \( \left(x^2 - \frac{x}{2} + \frac{1}{16}\right) + y^2 = \frac{1}{16} \)
or \( \left(x - \frac{1}{4}\right)^2 + y^2 = \frac{1}{16} \)
Therefore, the given circle has centre at \( \left(\frac{1}{4}, 0\right) \) and radius \( \frac{1}{4} \).

Example 2. 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.
Answer:
Let the equation of the circle be (x − h)² + (y − k)² = r²
Since the circle passes through (4, 1) and (6, 5), we have
(4 − h)² + (1 − k)² = r² ... (i)
and (6 − h)² + (5 − k)² = r² ... (ii)
Also, since the centre lies on the line 4x + y = 16, we have 4h + k = 16 ... (iii)
Simplifying equation (i), we get 16 − 8h + h² + 1 − 2k + k² = r², or 17 − 8h + h² − 2k + k² = r²
From equation (ii), we get 36 − 12h + h² + 25 − 10k + k² = r², or 61 − 12h + h² − 10k + k² = r²
Subtracting (eliminating the square terms): 17 − 61 − 8h + 12h − 2k + 10k = 0
or −44 + 4h + 8k = 0, or 4h + 8k = 44 ... (iv)
Solving equations (iii) and (iv): subtracting (iii) from (iv), 7k = 28 ⇒ k = 4
Substituting k = 4 in equation (iii), we get 4h + 4 = 16, or 4h = 12, or h = 3
Substituting h = 3, k = 4 in equation (i), we get (4 − 3)² + (1 − 4)² = r², or 1 + 9 = r², or r² = 10
Hence, the required equation of the circle is (x − 3)² + (y − 4)² = 10
or x² − 6x + 9 + y² − 8y + 16 = 10
or x² + y² − 6x − 8y + 15 = 0

Example 3. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).
Answer:
Let the equation of the circle be (x − h)² + (y − k)² = r²
Since the circle passes through (2, 3) and its radius is 5,
(2 − h)² + (3 − k)² = 25
or 4 − 4h + h² + 9 − 6k + k² = 25
or −12 − 4h + h² − 6k + k² = 0 ... (i)
Also, since the centre lies on the x-axis, we have k = 0 ... (ii)
Putting k = 0 in equation (i), we have −12 − 4h + h² = 0
or h² − 4h − 12 = 0
or h² − 6h + 2h − 12 = 0
or h(h − 6) + 2(h − 6) = 0
or (h − 6)(h + 2) = 0
or h = 6 or h = −2
Hence, the required equations of the circle are
(a) For h = 6: (x − 6)² + (y − 0)² = 5², or x² − 12x + 36 + y² = 25, or x² + y² − 12x + 11 = 0
(b) For h = −2: (x − (−2))² + (y − 0)² = 5², or (x + 2)² + y² = 25, or x² + 4x + 4 + y² = 25, or x² + y² + 4x − 21 = 0

Example 4. Find the equation of a circle with centre (2, 2) which passes through the point (4, 5).
Answer:
Since the centre is (2, 2), we have (x − 2)² + (y − 2)² = r²
This circle passes through the point (4, 5):
(4 − 2)² + (5 − 2)² = r²
or 4 + 9 = r², or r² = 13
The required equation of the circle is (x − 2)² + (y − 2)² = 13
or x² − 4x + 4 + y² − 4y + 4 = 13
or x² + y² − 4x − 4y − 5 = 0

Example 5. A circle has radius equal to 3 units and its centre lies on the line y = x − 1. Find the equation of the circle if it passes through (7, 3).
Answer:
Let the centre of the circle be (α, β). It lies on the line y = x − 1
⇒ β = α − 1. Hence the centre is (α, α − 1).
⇒ Then the equation of the circle is (x − α)² + (y − α + 1)² = 9.
It passes through (7, 3)
⇒ (7 − α)² + (4 − α)² = 9
⇒ α² − 11α + 28 = 0
⇒ (α − 4)(α − 7) = 0 ⇒ α = 4, 7.
Hence the required equations are x² + y² − 8x − 6y + 16 = 0 and x² + y² − 14x − 12y + 76 = 0.

Example 6. Find the equation of the circle that passes through the points (1, 0), (−1, 0) and (0, 1).
Answer:
Let the required circle be x² + y² + 2gx + 2fy + c = 0 ... (i)
It passes through (1, 0), (−1, 0) and (0, 1). Therefore, on substituting the coordinates of the three points successively in equation (i), we get
1 + 2g + c = 0 ... (ii)
1 − 2g + c = 0 ... (iii)
1 + 2f + c = 0 ... (iv)
Subtracting (iii) from (ii), we get 4g = 0 ⇒ g = 0
Putting g = 0 in (ii), we obtain c = −1
Now putting c = −1 in (iv), we get f = 0
Substituting the values of g, f and c in (i), we obtain the equation of the required circle as x² + y² = 1

Example 7. Find the equation of the circle which passes through the points (5, −8), (2, −9) and (2, 1). Find also the coordinates of its centre and radius.
Answer:
Let the required circle be x² + y² + 2gx + 2fy + c = 0 ... (i)
It passes through (5, −8), (2, −9) and (2, 1). Therefore,
89 + 10g − 16f + c = 0 ... (ii)
85 + 4g − 18f + c = 0 ... (iii)
5 + 4g + 2f + c = 0 ... (iv)
Subtracting (iii) from (ii), we obtain 4 + 6g + 2f = 0 ⇒ 2 + 3g + f = 0 ... (v)
Subtracting (iv) from (iii), we get 80 + 0g − 20f = 0 ⇒ f = 4
Putting f = 4 in (v), we get g = −2
Putting f = 4, g = −2 in (iv), we get 5 − 8 + 8 + c = 0 ⇒ c = −5
Substituting the values of g, f, c in (i), we obtain the equation of the required circle as x² + y² − 4x + 8y − 5 = 0
The coordinates of the centre are (−g, −f), i.e., (2, −4)
and, Radius = \( \sqrt{g^2 + f^2 - c} = \sqrt{4 + 16 + 5} = 5 \)

Example 8. The straight line \( \frac{x}{a} + \frac{y}{b} = 1 \) cuts the coordinate axes at A and B. Find the equation of the circle passing through O (0, 0), A and B.
Answer:
The straight line \( \frac{x}{a} + \frac{y}{b} = 1 \) cuts the coordinate axes at A (a, 0) and B (0, b).
Let x² + y² + 2gx + 2fy + c = 0 be the circle passing through O, A and B. Then, ... (i)
0 + 0 + c = 0 ... (ii)
a² + 2ga + c = 0 ... (iii)
b² + 2fb + c = 0 ... (iv)
Solving (ii), (iii) and (iv), we obtain \( g = -\frac{a}{2}, f = -\frac{b}{2} \) and c = 0
Substituting these values in (i), we obtain the equation of the required circle as x² + y² − ax − by = 0

Example 9. Find the equation of the circle passing through (1, 0) and (0, 1) and having the smallest possible radius.
Answer:
Let the equation of the required circle be x² + y² + 2gx + 2fy + c = 0 ... (i)
This passes through A (1, 0) and B (0, 1).
∴ 1 + 2g + c = 0 and 1 + 2f + c = 0
⇒ \( g = -\left(\frac{c + 1}{2}\right) \) and \( f = -\left(\frac{c + 1}{2}\right) \)
Let r be the radius of circle (i). Then, \( r = \sqrt{g^2 + f^2 - c} \)
⇒ \( r = \sqrt{\left(\frac{c + 1}{2}\right)^2 + \left(\frac{c + 1}{2}\right)^2 - c} \)
⇒ \( r = \sqrt{\frac{c^2 + 1}{2}} \)
⇒ \( r^2 = \frac{1}{2}(c^2 + 1) \)
Clearly, r is minimum when c = 0 and the minimum value of r is \( \frac{1}{\sqrt{2}} \).
For c = 0, we have \( g = -\frac{1}{2} \) and \( f = -\frac{1}{2} \)
Substituting the values of g, f and c in (i), we get x² + y² − x − y = 0 as the equation of the required circle.

Example 10. Show that the points (9, 1), (7, 9), (−2, 12) and (6, 10) are concyclic.
Answer:
Let the equation of the circle passing through (9, 1), (7, 9) and (−2, 12) be x² + y² + 2gx + 2fy + c = 0 ... (i)
Then, 82 + 18g + 2f + c = 0 ... (ii)
130 + 14g + 18f + c = 0 ... (iii)
148 − 4g + 24f + c = 0 ... (iv)
Subtracting (ii) from (iii), we get 48 − 4g + 16f = 0 ⇒ 12 − g + 4f = 0 ... (v)
Subtracting (iii) from (iv), we get 18 − 18g + 6f = 0 ⇒ 3 − 3g + f = 0 ... (vi)
Solving (v) and (vi) as simultaneous linear equations in g and f, we get f = −3, g = 0
Putting f = −3, g = 0 in (ii), we get 82 + 0 − 6 + c = 0 ⇒ c = −76
Substituting the values of g, f, c in (i), we get x² + y² − 6y − 76 = 0 as the equation of the circle passing through (9, 1), (7, 9) and (−2, 12).
Clearly, the point (6, 10) satisfies this equation (36 + 100 − 60 − 76 = 0). Hence, the given points are concyclic.

Example 11. Find the equation of the circle which passes through the points (1, −2) and (4, −3) and has its centre on the line 3x + 4y = 7.
Answer:
Let the equation of the required circle be x² + y² + 2gx + 2fy + c = 0 ... (i)
It passes through (1, −2) and (4, −3).
∴ 5 + 2g − 4f + c = 0 ... (ii)
and, 25 + 8g − 6f + c = 0 ... (iii)
The centre (−g, −f) of (i) lies on 3x + 4y = 7. Therefore, −3g − 4f = 7 ... (iv)
Subtracting (ii) from (iii), we get 20 + 6g − 2f = 0 ⇒ 10 + 3g − f = 0 ... (v)
Solving (iv) and (v) as simultaneous equations, we get \( g = -\frac{47}{15} \) and \( f = \frac{3}{5} \)
Substituting the values of g and f in (ii), we get \( 5 - \frac{94}{15} - \frac{12}{5} + c = 0 \Rightarrow c = \frac{55}{15} = \frac{11}{3} \)
Substituting the values of g, f and c in (i), we obtain the required equation of the circle as
\( x^2 + y^2 - \frac{94}{15}x + \frac{6}{5}y + \frac{11}{3} = 0 \)
⇒ 15(x² + y²) − 94x + 18y + 55 = 0

Example 12. Find the equation of the circle circumscribing the triangle formed by the lines x + y = 6, 2x + y = 4 and x + 2y = 5.
Answer:
Let the equations of sides AB, BC and CA of △ABC be respectively
x + y = 6 ... (i), 2x + y = 4 ... (ii) and x + 2y = 5 ... (iii)
Solving (i) and (iii); (i) and (ii); (ii) and (iii), we get the coordinates of A, B and C. The coordinates of A, B and C are (7, −1), (−2, 8) and (1, 2) respectively.
Let the equation of the circumcircle of △ABC be x² + y² + 2gx + 2fy + c = 0 ... (iv)
It passes through A (7, −1), B (−2, 8) and C (1, 2). Therefore,
50 + 14g − 2f + c = 0 ... (v)
68 − 4g + 16f + c = 0 ... (vi)
5 + 2g + 4f + c = 0 ... (vii)
Subtracting (v) from (vi), we get 18 − 18g + 18f = 0 ⇒ 1 − g + f = 0 ... (viii)
Subtracting (v) from (vii), we get −45 − 12g + 6f = 0 ... (ix)
Solving (viii) and (ix), we get g = −17/2, f = −19/2.
Putting the values of g and f in (v), we get c = 50.
Substituting the values of g, f and c in (iv), the equation of the required circumcircle is x² + y² − 17x − 19y + 50 = 0

Example 13. Find the radius of the circle (x cos α + y sin α − a)² + (x sin α − y cos α − b)² = k². If α varies, the locus of its centre is again a circle. Also, find its centre and radius.
Answer:
The given equation is (x cos α + y sin α − a)² + (x sin α − y cos α − b)² = k²
⇒ x²(cos²α + sin²α) + y²(sin²α + cos²α) − 2(a cos α + b sin α)x − 2(a sin α − b cos α)y + a² + b² − k² = 0
⇒ x² + y² − 2x(a cos α + b sin α) − 2y(a sin α − b cos α) + a² + b² − k² = 0
The coordinates of the centre of this circle are (a cos α + b sin α, a sin α − b cos α)
and, Radius = \( \sqrt{(a\cos\alpha + b\sin\alpha)^2 + (a\sin\alpha - b\cos\alpha)^2 - (a^2 + b^2 - k^2)} \)
⇒ Radius = \( \sqrt{a^2(\cos^2\alpha + \sin^2\alpha) + b^2(\sin^2\alpha + \cos^2\alpha) - (a^2 + b^2 - k^2)} \)
⇒ Radius = \( \sqrt{a^2 + b^2 - a^2 - b^2 + k^2} = k \)
Let (p, q) be the coordinates of the centre of the given circle. Then, p = a cos α + b sin α and q = a sin α − b cos α
To find the locus of (p, q) we have to eliminate α. Squaring and adding these two, we get
p² + q² = (a cos α + b sin α)² + (a sin α − b cos α)²
⇒ p² + q² = a²(cos²α + sin²α) + b²(sin²α + cos²α)
⇒ p² + q² = a² + b²
Hence, the locus of (p, q) is x² + y² = a² + b². This is a circle having centre at (0, 0) and radius equal to \( \sqrt{a^2 + b^2} \).

Example 14. Find the area of an equilateral triangle inscribed in the circle x² + y² + 2gx + 2fy + c = 0.
Answer:
Let ABC be an equilateral triangle inscribed in the circle x² + y² + 2gx + 2fy + c = 0 ... (i)
Let O (−g, −f) be the centre of the circle. Then, \( OA = OB = OC = \sqrt{g^2 + f^2 - c} \)

In △OBD (D is the mid-point of BC and ∠BOD = 60°), we have \( \sin 60^\circ = \frac{BD}{OB} \)
⇒ \( BD = \frac{\sqrt{3}}{2}\sqrt{g^2 + f^2 - c} \)
⇒ BC = 2 BD ⇒ \( BC = \sqrt{3}\sqrt{g^2 + f^2 - c} \)
∴ Area of △ABC = \( \frac{\sqrt{3}}{4} \)(Side)²
⇒ Area of △ABC = \( \frac{\sqrt{3}}{4} \times 3(g^2 + f^2 - c) = \frac{3\sqrt{3}}{4}(g^2 + f^2 - c) \) sq. units.

Example 15. Find the equation of the circle whose diameter is the line joining the points (−4, 3) and (12, −1). Find also the intercept made by it on y-axis.
Answer:
Equation of the circle having (−4, 3) and (12, −1) as the ends of a diameter is
(x + 4)(x − 12) + (y − 3)(y + 1) = 0
⇒ x² + y² − 8x − 2y − 51 = 0 ... (1)
Comparing (1) with the standard equation of circle x² + y² + 2gx + 2fy + c = 0,
g = −4, f = −1, c = −51
∴ Intercept on y-axis = \( 2\sqrt{f^2 - c} = 2\sqrt{1 + 51} = 4\sqrt{13} \).

Example 16. Find the equation of the circle which touches the axis of y at a distance of 4 units from the origin and cuts an intercept of 6 units from the axis of x.
Answer:

Let C be the centre, N the point where the circle touches the y-axis and PQ the intercept on the x-axis with mid-point M.
∵ CM = NO = 4 and PM = 3 (half of the intercept 6)
In △PCM, (PC)² = (3)² + (4)²
∴ PC = 5, i.e., radius of circle = 5
∴ NC = 5, so the centre of the circle is (5, 4)
∴ Equation of circle, if the centre is in the I quadrant: (x − 5)² + (y − 4)² = 25
If the centre is in the II, III and IV quadrants, then the equations are (x + 5)² + (y − 4)² = 25, (x + 5)² + (y + 4)² = 25 and (x − 5)² + (y + 4)² = 25
Hence, there are 4 circles which satisfy the given conditions. They are (x ± 5)² + (y ± 4)² = 25
or x² + y² ± 10x ± 8y + 16 = 0