# How to do this perpendicular bisector of a chord question?

## Unable to figure this out and the answer gives no explanation in the textbook. Could someone kindly help? Jan 16, 2017

The center of the circle is $= \left(3 , 3\right)$

#### Explanation:

The equation of a circle is

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

where $\left(a , b\right)$ is the centre of the circle

and $r$ is the radius

Plug in the equation of the circle for the 3 coordinates

Point A

${\left(5 - a\right)}^{2} + {\left(7 - b\right)}^{2} = {r}^{2}$

$25 - 10 a + {a}^{2} + 49 - 14 b + {b}^{2} = {r}^{2}$, $e q u a t i o n 1$

Point B

${\left(7 - a\right)}^{2} + {\left(1 - b\right)}^{2} = {r}^{2}$

$49 - 14 a + {a}^{2} + 1 - 2 b + {b}^{2} = {r}^{2}$, $e q u a t i o n 2$

Point C

${\left(- 1 - a\right)}^{2} + {\left(5 - b\right)}^{2} = {r}^{2}$

$1 + 2 a + {a}^{2} + 25 - 10 b + {b}^{2} = {r}^{2}$, $e q u a t i o n 3$

From equations 1 and 2, we get

$25 - 10 a + {\cancel{a}}^{2} + \cancel{49} - 14 b + {\cancel{b}}^{2} = \cancel{49} - 14 a + {\cancel{a}}^{2} + 1 - 2 b + {\cancel{b}}^{2}$

$4 a - 12 b = - 24$

$a - 3 b = - 6$, $e q u a t i o n 4$

From equation 2 and 3, we get

$49 - 14 a + {\cancel{a}}^{2} + 1 - 2 b + {\cancel{b}}^{2} = 1 + 2 a + {\cancel{a}}^{2} + 25 - 10 b + {\cancel{b}}^{2}$

$16 a - 8 b = 24$

$2 a - b = 3$ , $e q u a t i o n 5$

From equation $4$ and $5$, we get

$3 b - 6 = \frac{b + 3}{2}$

$6 b - 12 = b + 3$, $\implies$, $5 b = 15$, $\implies$, $b = 3$

$a = 3 b - 6 = 3 \cdot 3 - 6 = 3$

Therefore,

the centre of the circle is $= \left(3 , 3\right)$

The radius of the circle is

${r}^{2} = {\left(5 - 3\right)}^{2} + {\left(7 - 3\right)}^{2} = 4 + 16 = 20$

Jan 16, 2017

(3, 3 )

#### Explanation:

Here AB is a chord, o is the center of the circle. OD is the perpendicular of the chord.Three points A, B and C may be the points of a triangle ABC which is inscribed of a circle.we have to find out the co-ordinates of o.
Rest procedure as above done.

Jan 16, 2017

As explained below

#### Explanation:

Perpendicular bisector of a chord of a circle passes through the centre.

To prove this, draw a chord AB of a circle. If M is the midpoint of AB, the perpendicular bisector would passe through M. Now take any point P on this perpendicular bisector. Join A and B with P.
Now consider the right triangles APM and BPM. Both these triangles would be congruent by SAS, because AM=BM, PM=PM and angle AMP= angle BMP= ${90}^{o}$.
Therefore AP=BP. Since P is any point on the perpendicular bisector, it can be concluded that all points on the perpendicular bisector would be equidistant from A and B.

Now extend the perpendicular bisector on either side of chord AB, so that it meets the circle at points Q and R. Then AQ= BQ and AR=BR. The two triangles thus formed AQR and BQR would be congruent by SSS because AQ=BQ, AR=BR and QR=QR. Thus angle QAR = angleQBR. Since AQBR is a cyclic quadrilateral, angles QAR and QBR will be supplementary. These two angles would thus be ${90}^{o}$ each. That means the two halves of the circle are semicircles. Hence QR would be the diameter of the circle.

Thus it is proved that perpendicular bisector of the chord AB passes through the centre of the circle. 