Question

Find the centre and radius of the circle \(2x^2 + 2y^2 = 3x - 5y + 7\).

• A. \(\left(\dfrac{3}{4}, -\dfrac{5}{4}\right),\ \dfrac{3\sqrt{10}}{4}\)
• B. \(\left(\dfrac{1}{4}, \dfrac{2}{4}\right),\ \dfrac{\sqrt{10}}{4}\)
• C. \(\left(\dfrac{3}{4}, -\dfrac{1}{4}\right),\ \dfrac{5\sqrt{10}}{4}\)
• D. None of these

Hint:

To find the centre and radius of a circle:

Convert the equation to the form ((x-h)^2+(y-k)^2=r^2)
Centre is ((h,k))
Radius is (sqrtr^2)

Answer 1

The Correct Option is A

Step 1: Rewrite the given equation in standard form. \[ 2x^2 + 2y^2 = 3x - 5y + 7 \] Bring all terms to one side: \[ 2x^2 + 2y^2 - 3x + 5y - 7 = 0 \] Divide throughout by \(2\): \[ x^2 + y^2 - \frac{3}{2}x + \frac{5}{2}y - \frac{7}{2} = 0 \] Step 2: Group \(x\) and \(y\) terms and complete the squares. \[ (x^2 - \tfrac{3}{2}x) + (y^2 + \tfrac{5}{2}y) = \tfrac{7}{2} \] Complete the square: \[ x^2 - \tfrac{3}{2}x + \left(\tfrac{3}{4}\right)^2 + y^2 + \tfrac{5}{2}y + \left(\tfrac{5}{4}\right)^2 = \tfrac{7}{2} + \tfrac{9}{16} + \tfrac{25}{16} \] \[ (x - \tfrac{3}{4})^2 + (y + \tfrac{5}{4})^2 = \tfrac{7}{2} + \tfrac{34}{16} \] \[ (x - \tfrac{3}{4})^2 + (y + \tfrac{5}{4})^2 = \tfrac{45}{16} \] Step 3: Compare with standard form. \[ (x-h)^2 + (y-k)^2 = r^2 \] Thus, \[ h = \tfrac{3}{4},\quad k = -\tfrac{5}{4},\quad r^2 = \tfrac{45}{16} \] \[ r = \sqrt{\tfrac{45}{16}} = \frac{3\sqrt{5}}{4}\sqrt{2} = \frac{3\sqrt{10}}{4} \] Step 4: Final conclusion. Centre \(= \left(\dfrac{3}{4}, -\dfrac{5}{4}\right)\) Radius \(= \dfrac{3\sqrt{10}}{4}\)