Question

Write the general equation for the circle that passes through the points (1, 1), (1, 3), and (9, 2).

Answer 1

Step-by-step explanation:

As we know that

• The circle center is equidistant from all three points, the distance being the circle radius.

• Any point equidistant from two points must lie on the perpendicular bisector of the line segment which join those two points.

• Which is, on the line through the midpoint of the line segment, perpendicular to the line segment.

The perpendicular bisector of the line segment joining the points (1, 1) and (1, 3) will be:

                   $\:y=\:\frac{1+3}{2}=\frac{4}{2}=2$

The perpendicular bisector of the line segment joining the points (1, 3), and (9, 2) will be:

                   $x=\:\frac{1+9}{2}=\frac{10}{2}=5$

These intersect at the center of the circle (5, 2).

The distance between (1, 1) and (5, 2) will be:

$\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$\mathrm{The\:distance\:between\:}\left(1,\:1\right)\mathrm{\:and\:}\left(5,\:2\right)\mathrm{\:is\:}$

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

$=\sqrt{4^2+1}$

$=\sqrt{16+1}$

$=\sqrt{17}$

So the equation of the circle can be written as:

$\left(x-5\right)^2+\left(y-2\right)^2=17$

$x^2-10x+y^2+29-4y=17$

$x^2-10x+y^2-4y+12\:=0$

$x^2+y^2-10x-4y+12=0$