Question

Question #86d19

Answer 1

`"Equation of the circle": (x - 4)^(2) + (y - 2)^(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; `"Midpoint" = (frac(x_(1) + x_(2))(2), frac(y_(1) + y_(2))(2))`:

`Rightarrow "Centre" = (frac(7 + 1)(2), frac(- 3 + 7)(2))`

`Rightarrow "Centre" = (frac(8)(2), frac(4)(2))`

`therefore "Centre" = (4, 2)`

Then, let's find the length of the diameter using the formula `d = sqrt((x_(1) - x_(2))^(2) + (y_(1) - y_(2))^(2))`:

`Rightarrow "Length of diameter" = sqrt((1 - 7)^(2) + (7 - (- 3))^(2))`

`Rightarrow "Length of diameter" = sqrt((- 6)^(2) + (10)^(2))`

`Rightarrow "Length of diameter" = sqrt(36 + 100)`

`Rightarrow "Length of diameter" = sqrt(136)`

`therefore "Length of radius" = frac(sqrt(136))(2)`

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

`Rightarrow (x - 4)^(2) + (y - 2)^(2) = (frac(sqrt(136))(2))^(2)`

`Rightarrow (x - 4)^(2) + (y - 2)^(2) = frac(136)(4)`

`therefore (x - 4)^(2) + (y - 2)^(2) = 34`

Answer 2

`x^2+y^2-8x-4y-14=0.`

Explanation:

We know that, the eqn. of a circle having centre at point `C(h,k)`

and radius `r` is, `(x-h)^2+(y-k)^2=r^2.`

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

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

`((7+1)/2,(-3+7)/2)=C(4,2).`

Also, `P(1,7)` is on the circle. `:. CP^2=r^2.`

`:, r^2=(4-1)^2+(7-2)^2=34.`

Therefore, the eqn. of the circle is,

`(x-4)^2+(y-2)^2=34, i.e.,`

`x^2+y^2-8x-4y-14=0.`

N.B. :-

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

diametrically opposite points `(x_1,y_1) and (x_2,y_2) :`

`Eqn. : (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0.`

Enjoy Maths.!