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Notely Maths 12
Class 12 Maths
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6.1 Q-2

Question Statement

In this problem, we are given different forms of the equation of a circle and are asked to find the center and radius for each case. Specifically, we will address the following:

  • Part (a): Find the center and radius for the equation x2+y2+12x−10y=0x^2 + y^2 + 12x - 10y = 0.
  • Part (b): For the equation 5x2+5y2+14x+12y−10=05x^2 + 5y^2 + 14x + 12y - 10 = 0, find the center and radius.
  • Part (c): For the equation x2+y2+6x+4y+13=0x^2 + y^2 + 6x + 4y + 13 = 0, determine the center and radius.
  • Part (d): For the equation 4x2+4y2−8x+12y−25=04x^2 + 4y^2 - 8x + 12y - 25 = 0, find the center and radius.

Background and Explanation

The general equation of a circle is given by:

(x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2

Where:

  • (h,k)(h, k) is the center of the circle.
  • rr is the radius of the circle.

In some cases, we are provided with the expanded general form of the circle’s equation:

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

Where:

  • g=−hg = -h, f=−kf = -k, and c=h2+k2−r2c = h^2 + k^2 - r^2.

By comparing the given equation with the general form, we can extract the values for gg, ff, and cc, which will help us determine the center (−g,−f)(-g, -f) and the radius r=g2+f2−cr = \sqrt{g^2 + f^2 - c}.

Solution

Part (a)

Given: The equation x2+y2+12x−10y=0x^2 + y^2 + 12x - 10y = 0.

  • Step 1: Compare this with the general form of the circle equation x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0.

    • We can see that 2g=122g = 12 and 2f=−102f = -10, so: g=6andf=−5g = 6 \quad \text{and} \quad f = -5

  • Step 2: The constant term cc is 0.

  • Step 3: The center of the circle is: (−g,−f)=(−6,5)(-g, -f) = (-6, 5)

  • Step 4: To find the radius, use the formula: r=g2+f2−cr = \sqrt{g^2 + f^2 - c}
    Substituting the values: r=62+(−5)2−0=36+25=61r = \sqrt{6^2 + (-5)^2 - 0} = \sqrt{36 + 25} = \sqrt{61}

Thus, the center is (−6,5)(-6, 5) and the radius is 61\sqrt{61}.

Part (b)

Given: The equation 5x2+5y2+14x+12y−10=05x^2 + 5y^2 + 14x + 12y - 10 = 0.

  • Step 1: First, divide the entire equation by 5 to simplify: x2+y2+145x+125y−2=0x^2 + y^2 + \frac{14}{5}x + \frac{12}{5}y - 2 = 0

  • Step 2: Compare this with the general form of the circle equation x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0:

    • From the equation, we have: 2g=145⇒g=752g = \frac{14}{5} \quad \Rightarrow g = \frac{7}{5} 2f=125⇒f=652f = \frac{12}{5} \quad \Rightarrow f = \frac{6}{5} c=−2c = -2

  • Step 3: The center of the circle is: (−g,−f)=(−75,−65)(-g, -f) = \left( -\frac{7}{5}, -\frac{6}{5} \right)

  • Step 4: To find the radius, use the formula: r=g2+f2−cr = \sqrt{g^2 + f^2 - c}
    Substituting the values: r=(−75)2+(−65)2+2r = \sqrt{\left(\frac{-7}{5}\right)^2 + \left(\frac{-6}{5}\right)^2 + 2} r=4925+3625+2=49+36+5025=13525=275r = \sqrt{\frac{49}{25} + \frac{36}{25} + 2} = \sqrt{\frac{49 + 36 + 50}{25}} = \sqrt{\frac{135}{25}} = \sqrt{\frac{27}{5}}

Thus, the center is (−75,−65)\left( -\frac{7}{5}, -\frac{6}{5} \right) and the radius is 275\sqrt{\frac{27}{5}}.

Part (c)

Given: The equation x2+y2+6x+4y+13=0x^2 + y^2 + 6x + 4y + 13 = 0.

  • Step 1: Compare this with the general form of the circle equation x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0:

    • We have: 2g=6⇒g=32g = 6 \quad \Rightarrow g = 3 2f=4⇒f=22f = 4 \quad \Rightarrow f = 2 c=13c = 13

  • Step 2: The center of the circle is: (−g,−f)=(3,−2)(-g, -f) = (3, -2)

  • Step 3: To find the radius, use the formula: r=g2+f2−cr = \sqrt{g^2 + f^2 - c}
    Substituting the values: r=32+22−13=9+4−13=0r = \sqrt{3^2 + 2^2 - 13} = \sqrt{9 + 4 - 13} = \sqrt{0}

Thus, the center is (3,−2)(3, -2) and the radius is 00 (indicating a point at the center).

Part (d)

Given: The equation 4x2+4y2−8x+12y−25=04x^2 + 4y^2 - 8x + 12y - 25 = 0.

  • Step 1: Divide the entire equation by 4: x2+y2−2x+3y−254=0x^2 + y^2 - 2x + 3y - \frac{25}{4} = 0

  • Step 2: Compare this with the general form of the circle equation x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0:

    • We have: g=−1,f=−32,c=254g = -1, \quad f = -\frac{3}{2}, \quad c = \frac{25}{4}

  • Step 3: The center of the circle is: (−g,−f)=(1,−32)(-g, -f) = \left( 1, -\frac{3}{2} \right)

  • Step 4: To find the radius, use the formula: r=g2+f2−cr = \sqrt{g^2 + f^2 - c}
    Substituting the values: r=12+(−32)2+254r = \sqrt{1^2 + \left(\frac{-3}{2}\right)^2 + \frac{25}{4}} r=1+94+254=384=192r = \sqrt{1 + \frac{9}{4} + \frac{25}{4}} = \sqrt{\frac{38}{4}} = \sqrt{\frac{19}{2}}

Thus, the center is (1,−32)\left( 1, -\frac{3}{2} \right) and the radius is 192\sqrt{\frac{19}{2}}.

Key Formulas or Methods Used

  • General Equation of a Circle:
    x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0
    Where the center is (−g,−f)(-g, -f) and the radius is r=g2+f2−cr = \sqrt{g^2 + f^2 - c}.

  • Simplification of the Given Equation:
    Divide or adjust the equation to match the general form if necessary.

  • Radius Calculation:
    Using the formula r=g2+f2−cr = \sqrt{g^2 + f^2 - c}, calculate the radius after identifying gg, ff, and cc.

Summary of Steps

  1. Part (a): Compare the given equation to the general form and solve for gg, ff, and cc to find the center and radius.
  2. Part (b): Simplify the equation, compare with the general form, and find the center and radius.
  3. Part (c): Compare with the general form, calculate the center and radius (note: radius is 00 in this case).
  4. Part (d): Simplify the equation, find the center and radius, and express the result.