Question

The equation of the circle passing through $(4,6)$ and having centre $(1,2)$ is

• A. $x^2+y^2-2x-4y-20=0$
• B. $x^2+y^2-2x+4y-20=0$
• C. $x^2+y^2+2x-4y-20=0$
• D. $x^2+y^2+2x+4y-20=0$

Answer 1

The correct option is A $x^2+y^2-2x-4y-20=0$

Let coordinates of point A are $(4,6)$

As circle is passing through point A, point A lies on circle.

Let center of circle is $C(1,2)$

$\therefore h=1$ and $k=2$

Thus, AC is radius of circle.

By distance formula,

$AC=r=\sqrt{{(4-1)}^2+{(6-2)}^2}$

$\therefore r=\sqrt{{\left(3\right)}^2+{\left(4\right)}^2}$

$\therefore r=\sqrt{9+16}$

$\therefore r=\sqrt{25}$

$\therefore r=5$

Thus, equation of circle is,

${(x-h)}^2+{(y-k)}^2=r^2$

${(x-1)}^2+{(y-2)}^2={\left(5\right)}^2$

$x^2-2x+1+y^2-4y+4=25$

$\therefore x^2+y^2-2x-4y-20=0$

Thus, answer is option (A)