# Question #86d19

Jun 12, 2017

$\text{Equation of the circle} : {\left(x - 4\right)}^{2} + {\left(y - 2\right)}^{2} = \frac{136}{4}$

#### Explanation:

In order to determine the equation of a circle, we need to know its radius and the coordinates of its centre.

We are given the endpoints of a circle's diameter.

So let's find the coordinates of the centre using the midpoint formula; $\text{Midpoint} = \left(\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}\right)$:

$R i g h t a r r o w \text{Centre} = \left(\frac{7 + 1}{2} , \frac{- 3 + 7}{2}\right)$

$R i g h t a r r o w \text{Centre} = \left(\frac{8}{2} , \frac{4}{2}\right)$

$\therefore \text{Centre} = \left(4 , 2\right)$

Then, let's find the length of the diameter using the formula $d = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$:

$R i g h t a r r o w \text{Length of diameter} = \sqrt{{\left(1 - 7\right)}^{2} + {\left(7 - \left(- 3\right)\right)}^{2}}$

$R i g h t a r r o w \text{Length of diameter} = \sqrt{{\left(- 6\right)}^{2} + {\left(10\right)}^{2}}$

$R i g h t a r r o w \text{Length of diameter} = \sqrt{36 + 100}$

$R i g h t a r r o w \text{Length of diameter} = \sqrt{136}$

$\therefore \text{Length of radius} = \frac{\sqrt{136}}{2}$

Now, the equation of a circle is of the form ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$; where $\left(h , k\right)$ are the coordinates of the centre and $r$ is the radius:

$R i g h t a r r o w {\left(x - 4\right)}^{2} + {\left(y - 2\right)}^{2} = {\left(\frac{\sqrt{136}}{2}\right)}^{2}$

$R i g h t a r r o w {\left(x - 4\right)}^{2} + {\left(y - 2\right)}^{2} = \frac{136}{4}$

$\therefore {\left(x - 4\right)}^{2} + {\left(y - 2\right)}^{2} = 34$

Jun 12, 2017

${x}^{2} + {y}^{2} - 8 x - 4 y - 14 = 0.$

#### Explanation:

We know that, the eqn. of a circle having centre at point $C \left(h , k\right)$

and radius $r$ is, ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2.}$

Since the Mid-point of a Diameter is the Centre of the

circle, we see that, the centre $C$ is,

$\left(\frac{7 + 1}{2} , \frac{- 3 + 7}{2}\right) = C \left(4 , 2\right) .$

Also, $P \left(1 , 7\right)$ is on the circle. $\therefore C {P}^{2} = {r}^{2.}$

$: , {r}^{2} = {\left(4 - 1\right)}^{2} + {\left(7 - 2\right)}^{2} = 34.$

Therefore, the eqn. of the circle is,

${\left(x - 4\right)}^{2} + {\left(y - 2\right)}^{2} = 34 , i . e . ,$

${x}^{2} + {y}^{2} - 8 x - 4 y - 14 = 0.$

N.B. :-

We can use the following formula to find the eqn. of a circle having

diametrically opposite points $\left({x}_{1} , {y}_{1}\right) \mathmr{and} \left({x}_{2} , {y}_{2}\right) :$

$E q n . : \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) + \left(y - {y}_{1}\right) \left(y - {y}_{2}\right) = 0.$

Enjoy Maths.!