Question

The length of the diameter of the circle ${x^2} + {y^2} - 4x - 6y + 4 = 0$
A) 9
B) 3
C) 4
D) 6

Answer 1

Hint:
We will compare this given equation of circle with the standard equation of the circle given by: ${x^2} + {y^2} + 2gx + 2fy + c = 0$ . Then, we will calculate the centre of the circle as $(-g, -f)$ and hence, we will find the radius of the circle given by the formula: $r = \sqrt {{g^2} + {f^2} - c}$. The diameter of the circle is twice the radius of the circle and hence by putting the value of r, we will get the length of the diameter.

Complete step by step solution:
We are given the equation of the circle as: ${x}^{2} + {y}^{2} - 4x - 6y + 4 = 0$
Now, the standard equation of the circle is ${x}^{2} + {y}^{2} + 2gx + 2fy + c = 0$
Comparing the given equation with the standard equation, we get
$ \Rightarrow 2g = - 4, 2f = - 6$ and $c = 4$
$ \Rightarrow g = - 2, f = - 3$ and $c = 4$
Now, we know that the centre of the circle is at $(-g, -f)$
$ \Rightarrow $ Centre $ \equiv (2, 3)$
Now, the radius of the circle is given by the formula: $r = \sqrt {{g^2} + {f^2} - c} $.
Substituting the values of g, f and c in the equation of radius of the circle, we get
$ \Rightarrow $r = $\sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 3} \right)}^2} - \left( 4 \right)} $
$ \Rightarrow $r = $\sqrt {4 + 9 - 4} = \sqrt 9 = 3$
Therefore, the radius of the given circle is 3 units.
Now, the diameter of the circle is twice the radius of the circle.
$ \Rightarrow d = 2r = 2 (3) = 6$units.

Hence, option (D) is correct.

Note:
One alternative solution of the given problem is to convert the given equation of circle in the form of $\left(x-a\right)^{2}+\left(y-b\right)^{2}={r}^{2}$ where, $(a, b)$ is the centre and $r$ is the radius. So, just after converting the equation into this form we just need to take the square root of the right hand side number.