Question

The value of k, such that the equation ${\text{2}}{{\text{x}}^2} + 2{{\text{y}}^2} - 6{\text{x + 8y + k = 0}}$ represents a point circle, is
$
  {\text{A}}{\text{. 0}} \\
  {\text{B}}{\text{. 25}} \\
  {\text{C}}{\text{. }}\dfrac{{25}}{2} \\
  {\text{D}}{\text{. - }}\dfrac{{25}}{2} \\
$

Answer 1

Hint: In order to determine the value of k, we rewrite the given equation in the form of the equation of the circle. Also, for a point circle the radius is equal to zero.

Complete step-by-step answer:

Given, ${\text{2}}{{\text{x}}^2} + 2{{\text{y}}^2} - 6{\text{x + 8y + k = 0}}$ represents a point circle.
The equation of a circle: ${\left( {{\text{x - a}}} \right)^2} + {\left( {{\text{y - b}}} \right)^2} = {{\text{r}}^2}$
Now we write the given equation in this form.
Divide ${\text{2}}{{\text{x}}^2} + 2{{\text{y}}^2} - 6{\text{x + 8y + k = 0}}$ by 2
⟹${{\text{x}}^2} + {{\text{y}}^2} - 3{\text{x + 4y + }}\dfrac{{\text{k}}}{2} = 0$
Now we rearrange this equation, also add and subtract 4 and $\dfrac{9}{4}$to convert this equation in the form of the equation of circle.
⟹${{\text{x}}^2} - {\text{3x + }}\dfrac{9}{4} + {{\text{y}}^2} + 4{\text{y + 4}} - \dfrac{{ - 9}}{4} - 4 + \dfrac{{\text{k}}}{2} = 0$
⟹${\left( {{\text{x - }}\dfrac{3}{2}} \right)^2} + {\left( {{\text{y + 2}}} \right)^2} = \dfrac{{25}}{4} - \dfrac{{\text{k}}}{2}$
Now, for a point circle radius is equal to zero. On comparing radius terms in both the equations we get,
$\dfrac{{25}}{4} - \dfrac{{\text{k}}}{2}$= 0
⟹k = $\dfrac{{50}}{4}$=$\dfrac{{25}}{2}$
Hence, option C is the correct answer.

Note: In order to solve this type of problems the key is to be able to rewrite the given equation in terms of the equation of a circle. We then equate the radius term to 0, to find the answer. A point circle is just a point, a degenerate circle.