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Conic Sections - Circle

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

A circle is defined as the set of all points in a plane that are at a fixed distance, known as the radius rr, from a fixed point, known as the center (h,k)(h, k). Visually, the circle forms a perfectly round closed loop on the Cartesian plane where every point on the perimeter is equidistant from the midpoint.

The standard equation of a circle is (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2. This formula is derived from the distance formula; geometrically, it represents the locus of a point P(x,y)P(x, y) such that its distance from the center C(h,k)C(h, k) is always rr.

When the center of the circle is at the origin (0,0)(0, 0), the equation simplifies to x2+y2=r2x^2 + y^2 = r^2. Visually, the circle is centered at the intersection of the xx and yy axes, showing symmetry across both axes.

The general equation of a circle is expressed as x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0. For an equation to represent a circle, the coefficients of x2x^2 and y2y^2 must be equal and there should be no xyxy term. Geometrically, the center of this circle is located at (g,f)(-g, -f).

To find the radius rr from the general form, we use the expression g2+f2c\sqrt{g^2 + f^2 - c}. Visually, for a circle to be 'real', the value inside the square root (g2+f2cg^2 + f^2 - c) must be positive. If it is zero, the circle collapses into a single point (point circle).

A circle touching the axes has special geometric properties. If a circle touches the xx-axis, its radius is equal to the absolute value of the yy-coordinate of its center (r=kr = |k|). If it touches the yy-axis, the radius is equal to the absolute value of the xx-coordinate (r=hr = |h|).

The position of a point (x1,y1)(x_1, y_1) relative to a circle x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0 is determined by the sign of S1=x12+y12+2gx1+2fy1+cS_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c. Visually, if S1<0S_1 < 0, the point is inside the circle; if S1=0S_1 = 0, it is on the boundary; and if S1>0S_1 > 0, it lies outside.

📐Formulae

Standard Equation: (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2

Circle with Center at Origin: x2+y2=r2x^2 + y^2 = r^2

General Equation: x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0

Center of General Equation: C=(g,f)C = (-g, -f) where g=12(coeff. of x)g = \frac{1}{2}(\text{coeff. of } x) and f=12(coeff. of y)f = \frac{1}{2}(\text{coeff. of } y)

Radius of General Equation: r=g2+f2cr = \sqrt{g^2 + f^2 - c}

Condition for Real Circle: g2+f2c>0g^2 + f^2 - c > 0

Diameter Form of Circle: (xx1)(xx2)+(yy1)(yy2)=0(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 (where (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are endpoints of the diameter)

💡Examples

Problem 1:

Find the equation of the circle whose center is (2,3)(2, -3) and which passes through the point (5,1)(5, 1).

Solution:

  1. The center (h,k)(h, k) is given as (2,3)(2, -3).
  2. The radius rr is the distance between the center (2,3)(2, -3) and the point (5,1)(5, 1). Using the distance formula: r=(52)2+(1(3))2r = \sqrt{(5 - 2)^2 + (1 - (-3))^2} r=32+42=9+16=25=5r = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
  3. Substitute h=2h = 2, k=3k = -3, and r=5r = 5 into the standard equation (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2: (x2)2+(y(3))2=52(x - 2)^2 + (y - (-3))^2 = 5^2 (x2)2+(y+3)2=25(x - 2)^2 + (y + 3)^2 = 25
  4. Expanding the equation: x24x+4+y2+6y+9=25x^2 - 4x + 4 + y^2 + 6y + 9 = 25 x2+y24x+6y12=0x^2 + y^2 - 4x + 6y - 12 = 0

Explanation:

To find the equation, we first determine the radius by calculating the distance between the center and the given point on the circumference. We then use the center-radius form and expand it to get the general form.

Problem 2:

Find the center and radius of the circle represented by the equation 2x2+2y28x+12y6=02x^2 + 2y^2 - 8x + 12y - 6 = 0.

Solution:

  1. Divide the entire equation by 2 to make the coefficients of x2x^2 and y2y^2 equal to 1: x2+y24x+6y3=0x^2 + y^2 - 4x + 6y - 3 = 0
  2. Compare this with the general form x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0: 2g=4    g=22g = -4 \implies g = -2 2f=6    f=32f = 6 \implies f = 3 c=3c = -3
  3. The center is given by (g,f)(-g, -f): Center=((2),(3))=(2,3)\text{Center} = (-(-2), -(3)) = (2, -3)
  4. The radius rr is calculated as: r=g2+f2cr = \sqrt{g^2 + f^2 - c} r=(2)2+(3)2(3)r = \sqrt{(-2)^2 + (3)^2 - (-3)} r=4+9+3=16=4r = \sqrt{4 + 9 + 3} = \sqrt{16} = 4

Explanation:

Before identifying gg and ff, the equation must be normalized so that the coefficients of x2x^2 and y2y^2 are 1. Once in general form, we extract the center coordinates and apply the radius formula.