Question #86d19

2 Answers
Jun 12, 2017

#"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#

Jun 12, 2017

# 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.!