The length of the diameter of the circle $x^{2} + y^{2} - 4 x - 6 y + 4 = 0$ is -

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Solution

Given circle equation is:
$x^{2} + y^{2} - 4 x - 6 y + 4 = 0 . . . . . . . . . (1)$
Since we know that general equation of circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0....... (2)$ and radius of circle is $\sqrt{g^{2} + f^{2} - c}$ .
Now, compairing euation 1 and 2.
$g = - 2, f = - 3, & c = 4$ .
Radius= $\sqrt{(- 2)^{2} + (- 3)^{2} - 4}$
Radius= $\sqrt{9}$
Radius of given circle is $3$ .

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