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What do we mean by the locus of a point?
What is a Circle?
Perimeter of the Circle
Equation of a circle under Different Conditions
Parametric Equation of a circle
Diameter of circle
General equation of a circle in polar co-ordinate system
Particular cases of the general equation in polar coordinates
Director Circle
The Position of a Point with respect to a Circle
Related Resources
Locus of a point is defined to be the path of a point satisfying some geometrical condition; i.e. constraint equations. The path represents a curve, which includes all the points satisfying the given condition.
A circle is defined as the locus of a point which moves in such a way that its distance from a fixed point is always constant and positive. The fixed point is called the centre of the circle and the given distance the radius of the circle.
For eg: In real life, when you rotate a stone tied with one end of a string then the path followed by stone is exactly a circle whose centre is your finger an radius is length of the string.
The equation of a circle with its centre at C(x_{c}, y_{c}) and radius r is:
(x – x_{c})^{2} + (y – y_{c})^{2} = r^{2}
Proof:
Let P(x, y) be any point on the circle. Then by the definition of the locus the constant distance is (see figure given below)
|PC| = r ⇒ √(x-x_{c})^{2 }+ (y-y_{c}))^{2}= r ⇒ (x – x_{c})^{2} + (y – y_{c})^{2} = r^{2}
which is the required equation of the circle.
Remark: (1) If x_{c} = y_{c} = 0 (i.e. the centre of the circle is at origin) then equation of the circle reduce to x^{2} + y^{2} = r^{2}.
(2) If r = 0 then the circle represents a point or a point circle.
The perimeter of any figure refers to the sum of the boundaries. Similarly, in case of a circle, the perimeter is given by the circumference of the circle.
Equation of the circle in Various Forms
(i) The simplest equations of the circle is x^{2} + y^{2} = r^{2} whose centre is (0, 0) and radius ‘r’.
(ii) The equation (x – a)^{2}+ (y – b)^{2} = r^{2} represents a circle with centre (a, b) and radius r.
(iii) The equation x^{2} + y^{2} + 2gx + 2fy + c = 0 is the general equation of a circle with centre (–g, –f) and radius √(g^{2}+f^{2}-c).
(iv) Equation of the circle with points P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) as extremities of a diameter is (x – x_{1}) (x – x_{2}) + (y – y_{1})(y – y_{2}) = 0.
Let us consider a circle of radius ‘r’ and centre at C(x_{c}, y_{c}) we have:
(y-y_{c})/r = sin θ (see figure given below)
⇒ y = y_{c} + r sin θ
Similarly x = x_{c} + r cos θ
This gives the parametric form of the equation of a circle.
The locus of middle points of a system of parallel chords of a circle is called the diameter of the circle. The diameter of the circle x^{2} + y^{2} = r^{2} corresponding to the system of parallel chords y = mx + c is x + my = 0.
(a) Every diameter passes through the centre of the circle.
(b) A diameter is perpendicular to the system of parallel chords.
Let O be the origin, or pole, OX the initial line, C the centre and ‘a’ the radius of the circle.
Let the polar co-ordinates of C be R and α, so that OC = R and ∠XOC = α.
Let a radius vector through O at an angle θ with the initial line cut the circle at P and Q. Let OP be r.
Then we have
CP^{2} = OC^{2} + OP^{2} – 2OC . OP cos COP
i.e. a^{2} = R^{2} + r^{2} – 2 Rr cos (θ – α)
i.e. r^{2} – 2 Rr cos (θ – α) + R^{2} – a^{2} = 0 …… (1)
This is the required polar equation.
1. Let the initial line be taken to go through the centre C. Then α = 0, and the equation becomes
r^{2} – 2Rr cos θ + R^{2} – a^{2} = 0.
2. Let the pole O be taken on the circle, so that R = OC = α
The general equation the becomes
r^{2} – 2ar cos (θ – α) = 0,
i.e. r = 2a cos (θ – α).
3. Let the pole be on the circle and also let the initial line pass through the centre of the circle. In this case
α = 0, and R = a
Now, the general equation reduces to the simple form r = 2a cos θ
This is at once evident from the figure given above.
For, if OCA were a diameter, we have
OP = OA cos θ,
r = 2a cos θ.
∠PRQ = π/2 (Angle subtended by diameter at any point on the circle is a right angle).
⇒ QR ⊥ PR
⇒ (Slope of QR) x (Slope of PR) = –1
⇒ (y-y_{2})/(x-x_{2}) ×(y-y_{1})/(x-x_{1}) = – 1
⇒ (x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0
which gives the required equation.
This equation can also be obtained considering
PR^{2} + QR^{2} = PQ^{2} The general from of the equation of a circle is: x^{2} + y^{2} + 2gx + 2fy + c = 0 …… (1) ⇒ (x + g)^{2} + (y + f)^{2} = g^{2} + f^{2} – c Comparing this equation with the standard equation (x – x_{c})^{2} + (y – y_{c})^{2}= r^{2}
We have:
Centre of the circle is (–g, –f), Radius = √(g^{2}+f^{2}-c). Equation (1) is also written as S = 0.
1. If g^{2} + f^{2} – c > 0, circle is real
2. If g^{2} + f^{2} – c = 0, circle is a point circle.
3. If g^{2} + f^{2} – c < 0, the circle is imaginary.
4. Any second-degree equation ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 represents a circle only when h = 0 and a = b i.e. if there is no term containing xy and co-efficient of x^{2} and y^{2} are same, provided abc + 2fgh – af^{2} – bg^{2} – ch^{2} ≠ 0
The locus of the point of intersection of two perpendicular tangents to a circle is called the director circle.
We now try to find out the equation of director circle:
If the equation of the circle is x^{2} + y^{2} = a^{2}, then the equation of pair of tangents to a circle from a point (x_{1}, y_{1}) is
(x^{2} + y^{2} –a^{2}) (x_{1}^{2} + y_{1}^{2} – a^{2}) = (xx_{1} + yy_{1} – a^{2})^{2}
If this represents a pair of perpendicular lines then the coefficient of x^{2} + coefficient of y^{2} = 0
i.e. (x_{1}^{2} + y_{1}^{2} – a^{2} - x_{1}^{2}) + (x_{1}^{2 }+ y_{1}^{2} – a^{2} - y_{1}^{2}) = 0
This gives x_{1}^{2 }+ y_{1}^{2 }= 2a^{2}
Hence, this means that the equation of director circle is x^{2} + y^{2} = 2a^{2}
1. Director Circle is a concentric circle whose radius is √2 times the radius of the given circle.
2. The director circle of circle x^{2} + y^{2} + 2gx + 2fy + c = 0 is x^{2} + y^{2} + 2gx + 2fy + 2c - g^{2} - f^{2} = 0.
Find the centre and the radius 3x^{2} + 3y^{2} – 8x – 10y + 3 = 0.
We write the given equation as x^{2} + y^{2} – 8/3 x – 10/3 y + 1 = 0.
⇒ g = -4/3, f = -5/3 , c = 1
Hence the centre is (4/3,5/3) and the radius is
√(16/9+25/9-1)=√(32/9)=(4√2)/3.
Find the length of intercept on y-axis, by a circle whose diameter is the line joining the points (-4, 3) and (12, -1).
The equation of the required circle is (x+4)(x-12) + (y-3)(y+1) = 0
This gives the eqaution as x^{2} + y^{2} – 8x – 2y -51= 0.
Hence intercept on y-axis is 2√f^{2}-c = 2√(1-(-51)) = 4√13.
A circle has radius 3 units and its centre lies on the line y = x – 1. Find the equation of the circle if it passes through (7, 3).
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).
⇒ 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 ⇒ 2α^{2} – 22α + 56 = 0
⇒ α^{2} – 11α + 28 = 0 ⇒ (α – 7) = 0 ⇒ α = 4, 7.
Hence the required equations are
x^{2} + y^{2} – 8x – 6y + 6 = 0 and x^{2} + y^{2} – 14x – 12y + 76 = 0.
Illustration:
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 the y-axis.
The equation of the required circle is
(x + 4) (x – 12) + (y – 3) (y + 1) = 0.
On the y-axis, x = 0 ⇒ – 48 + y^{2} – 2y – 3 = 0.
⇒ y^{2} – 2y – 51 = 0 ⇒ y = 1 ± √52.
Hence the intercept on the y-axis = 22√52 = 4√13.
Find the equation of the circle passing through (1, 1), (2, –1) and (3, 2).
Let the equation be x^{2} + y^{2} + 2gx + 2fy + c = 0.
Substituting the coordinates of three points, we get 2g + 2f + c = –2, 4g – 2f + c = –5, 6g + 4f + c = –13. Solving the above three equations, we obtain: f = –1/2; g = –5/2, c = 4. Hence the equation of the circle is x^{2} + y^{2} – 5x – y + 4 = 0.
Write general equation of a circle centered at a point on x-axis.
Circle is: x^{2} + y^{2} + 2gx + c = 0, g^{2} – c ≥ 0 Its centre is (–g, 0) and radius √(g^{2}-c) Or (x + g)^{2} + (y – 0)^{2} = r^{2}
Its centre is (–g, 0) and radius r. (figure given above)
Write the equation of a circle passing through O (0, 0) A (a, 0) and B (0, b)? Obviously AB is the diameter of the circle. (Figure given below)
Since the circle passes through O (0, 0) A (a, 0) and B (0, b), hence we have the equation as
(x – a) (x – 0) + (y – 0) (y – b) = 0
Find the equation of circle shown in figure given below in polar form.
Clearly, OC is the radius. Hence, we have OP = OA cos θ.
This means r = 2a cos θ, – θ/2 ≤ θ ≤ θ/2, where a is radius of circle.
Find the co-ordinates of the centre of the circle represented by r = A cos θ + B sin θ.
r = A cos θ + B sin θ
= [A/√(A^{2}+B^{2} ) cos θ +B/√(A^{2}+B^{2} ) sin θ ] √(A^{2}+B^{2} )
= cos (θ – α) (√(A^{2}+B^{2} ))
centre is ≡ (1/2 √(A^{2}+B^{2} ),tan_{-1} (B/A) )
1. The equation of the circle through three non-collinear points
2. The circle x^{2} + y^{2} + 2gx + 2fy + c = 0 makes an intercept on x-axis if x^{2} + 2gx + c = 0 has real roots i.e. if g^{2} > c. And, the magnitude of the intercept is 2√(g^{2}-c).
The point P(x_{1}, y_{1}) lies outside, on, or inside a circle S ≡ x^{2} + y^{2} + 2gx + 2fy + c = 0, according as S_{1} ≡ x_{1}^{2} + y_{1}^{2} + 2gx_{1} + 2fy_{1} + c is greater than, equal to or less than 0.
S_{1} > 0 ⇒ Point is outside the circle
S_{1} = 0 ⇒ Point is on the circle
S_{1} < 0 ⇒ Point is inside the circle
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