Question

If ${x^2} + hxy + {y^2} - 6x - 2y + k = 0$ is the equation of the circle and $2$ is the radius of the circle, then find the values of $h$ and $k$?
(A) $h = 0,k = - 6$
(B) $h = 0,k = 6$
(C) $h = - 3,k = 6$
(D) $h = 3,k = 6$

Answer 1

Hint: The general equation of a circle is ${(x - h)^2} + {(y - k)^2} = {r^2}$ or we can expand this equation and write it as ${x^2} + {y^2} + Ax + By + C = 0$; here from the equation itself we can get the coordinates of the center and the radius of the circle. We know that the coordinates of the center is $(h,k)$ and the radius of the given circle is $r$.
We also know that the formula of radius of a circle is $r = \sqrt {{g^2} + {f^2} - c} $; where $( - g, - f)$ are the coordinates of the center of the given circle.

Complete step by step answer:
1. Let us first note down the given equation:
  ${x^2} + hxy + {y^2} - 6x - 2y + k = 0$
2. From the given equation we can see that it is of ${2^{nd}}$degree. But a ${2^{nd}}$degree equation can be written in a particular general form. Let us see how a second degree equation can be written in its respective general form;
$a{x^2} + b{y^2} + 2hxy + 2gx + 2fy + c = 0$
3. If I compare the given equation and the previous general equation then I can easily say that $a = b$and as per the condition of a circle if $a = b \Rightarrow h = 0$as coefficient of $xy$must be$0$.

4. Since we have made the coefficient of$xy = 0$, we can write the above given equations as:
${x^2} + {y^2} - 6x - 2y + k = 0$ and ${x^2} + {y^2} + 2gx + 2fy + c = 0$
5. We can compare the equations from step $4$and tell that:
Center: $( - g, - f) = (3,1)$
4. But as per the question, it is also given that radius of the circle is $2$and putting $r$and $c$in the formula of radius we can write:
$r = \sqrt {{g^2} + {f^2} - c} $
Now we can substitute the values of $r,g,f$in this equation:
$
   \Rightarrow {r^2} = {g^2} + {f^2} - c \\
   \Rightarrow 4 = 9 + 1 - c \\
   \Rightarrow c = 6 \\
 $
5. We also know that $k = c$by comparing the previous equations, hence the value of $k = 6$.
6. We have found out the values of $h$and $k$; that is $(h,k) = (0,6)$; option (B) is the right answer.

Note: A circle may be just a closed curve which is simple, but it has an equation. An equation of a circle is the algebraic form of describing a circle. The equation of a circle is able to tell us where exactly the center of the circle is and how much its radius is. As in the equation each term is derived based on the coordinates of the center and its distance from the center, every term in a circle’s equation allows us to find out the center and radius.