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Conic Sections - Circle

Grade 11CBSE

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

๐Ÿ”‘Concepts

โ€ข

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.

โ€ข

The standard equation of a circle is (xโˆ’h)2+(yโˆ’k)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.

โ€ข

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.

โ€ข

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 (โˆ’g,โˆ’f)(-g, -f).

โ€ข

To find the radius rr from the general form, we use the expression g2+f2โˆ’c\sqrt{g^2 + f^2 - c}. Visually, for a circle to be 'real', the value inside the square root (g2+f2โˆ’cg^2 + f^2 - c) must be positive. If it is zero, the circle collapses into a single point (point circle).

โ€ข

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=โˆฃkโˆฃr = |k|). If it touches the yy-axis, the radius is equal to the absolute value of the xx-coordinate (r=โˆฃhโˆฃr = |h|).

โ€ข

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.

๐Ÿ“Formulae

Standard Equation: (xโˆ’h)2+(yโˆ’k)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=(โˆ’g,โˆ’f)C = (-g, -f) where g=12(coeff.ย ofย x)g = \frac{1}{2}(\text{coeff. of } x) and f=12(coeff.ย ofย y)f = \frac{1}{2}(\text{coeff. of } y)

Radius of General Equation: r=g2+f2โˆ’cr = \sqrt{g^2 + f^2 - c}

Condition for Real Circle: g2+f2โˆ’c>0g^2 + f^2 - c > 0

Diameter Form of Circle: (xโˆ’x1)(xโˆ’x2)+(yโˆ’y1)(yโˆ’y2)=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)

๐Ÿ’กExamples

Problem 1:

Find the equation of the circle whose center is (2,โˆ’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,โˆ’3)(2, -3).
  2. The radius rr is the distance between the center (2,โˆ’3)(2, -3) and the point (5,1)(5, 1). Using the distance formula: r=(5โˆ’2)2+(1โˆ’(โˆ’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=โˆ’3k = -3, and r=5r = 5 into the standard equation (xโˆ’h)2+(yโˆ’k)2=r2(x - h)^2 + (y - k)^2 = r^2: (xโˆ’2)2+(yโˆ’(โˆ’3))2=52(x - 2)^2 + (y - (-3))^2 = 5^2 (xโˆ’2)2+(y+3)2=25(x - 2)^2 + (y + 3)^2 = 25
  4. Expanding the equation: x2โˆ’4x+4+y2+6y+9=25x^2 - 4x + 4 + y^2 + 6y + 9 = 25 x2+y2โˆ’4x+6yโˆ’12=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+2y2โˆ’8x+12yโˆ’6=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+y2โˆ’4x+6yโˆ’3=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=โˆ’4โ€…โ€ŠโŸนโ€…โ€Šg=โˆ’22g = -4 \implies g = -2 2f=6โ€…โ€ŠโŸนโ€…โ€Šf=32f = 6 \implies f = 3 c=โˆ’3c = -3
  3. The center is given by (โˆ’g,โˆ’f)(-g, -f): Center=(โˆ’(โˆ’2),โˆ’(3))=(2,โˆ’3)\text{Center} = (-(-2), -(3)) = (2, -3)
  4. The radius rr is calculated as: r=g2+f2โˆ’cr = \sqrt{g^2 + f^2 - c} r=(โˆ’2)2+(3)2โˆ’(โˆ’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.