Question 69 of 70intermediate🔧 ApplyShort Answer1 mark

In each of the following Exercises 6 to 9, find the centre and radius of the circles 2x² + 2y² - x = 0

Correct Answer

The given equation of the circle is 2x2+2y2−x=02x^2 + 2y^2 - x = 0.

Step 1: Divide the entire equation by 2 to make the coefficients of x2x^2 and y2y^2 equal to 1. x2+y2−x2=0x^2 + y^2 - \frac{x}{2} = 0

Step 2: Rearrange the terms to group the x terms together. (x2−x2)+y2=0(x^2 - \frac{x}{2}) + y^2 = 0

Step 3: Complete the square for the x-terms. Add (12×−12)2=116(\frac{1}{2} \times -\frac{1}{2})^2 = \frac{1}{16} to both sides. (x2−x2+116)+y2=116(x^2 - \frac{x}{2} + \frac{1}{16}) + y^2 = \frac{1}{16}

Step 4: Write the equation in the standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2. (x−14)2+(y−0)2=(14)2(x - \frac{1}{4})^2 + (y - 0)^2 = (\frac{1}{4})^2

Step 5: Compare with the standard form to find the centre and radius. Centre =(14,0)= (\frac{1}{4}, 0) Radius =14= \frac{1}{4}

Exercise: EXERCISE 10.1 | Q: 9 | (Chapter: 6)
For More Understanding

Explanation

To find the centre and radius, the equation must be converted to the standard form (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2. Following the method shown in Example 3 of the provided context, we first ensure the coefficients of x2x^2 and y2y^2 are unity. Then, we complete the square for the variable terms. By comparing the resulting equation with the standard form, the coordinates of the centre (h,k)(h, k) and the value of the radius rr are determined.

Solution Steps

  1. Divide the equation 2x2+2y2−x=02x^2 + 2y^2 - x = 0 by 2 to get x2+y2−x2=0x^2 + y^2 - \frac{x}{2} = 0.

  2. Rearrange terms: (x2−x2)+y2=0(x^2 - \frac{x}{2}) + y^2 = 0.

  3. Complete the square: (x2−x2+116)+y2=116(x^2 - \frac{x}{2} + \frac{1}{16}) + y^2 = \frac{1}{16}.

  4. Rewrite as perfect squares: (x−14)2+(y−0)2=(14)2(x - \frac{1}{4})^2 + (y - 0)^2 = (\frac{1}{4})^2.

  5. Identify centre as (14,0)(\frac{1}{4}, 0) and radius as 14\frac{1}{4}.

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