What is the standard form of the equation of a circle with centre is at point (5,8) and which passes through the point (2,5)?

2 Answers
Jan 5, 2016

Answer:

#(x - 5 )^2 + (y - 8 )^2 = 18 #

Explanation:

standard form of a circle is # (x - a )^2 + (y - b )^2 = r^2 #

where (a , b ) is the centre of the circle and r = radius.

in this question the centre is known but r is not. To find r , however ,

the distance from the centre to the point ( 2 , 5 ) is the radius . Using

the distance formula will allow us to find in fact # r^2 #

# r^2 = (x_2 - x_1 )^2 + ( y_2 - y_1 )^2 #

now using (2 , 5 ) = # (x_2 , y_2 ) and (5 , 8 ) = ( x_1 , y_1 ) #

then # (5 - 2 )^2 + ( 8 - 5 )^2 = 3^2 + 3^2 = 9 + 9 = 18 #

equation of circle : # (x - 5 )^2+ (y - 8 )^2 = 18 #
.

Jan 5, 2016

Answer:

I found: #x^2+y^2-10x-16y+71=0#

Explanation:

The distance #d# between the centre and the given point will be the radius #r#.
We can evaluate it using:
#d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#
So:
#r=d=sqrt((2-5)^2+(5-8)^2)=sqrt(9+9)=3sqrt(2)#
Now you can use the general form of the equation of a circle with centre at #(h,k)# and radius #r#:
#(x-h)^2+(y-k)^2=r^2#
And:
#(x-5)^2+(y-8)^2=(3sqrt(2))^2#
#x^2-10x+25+y^2-16y+64=18#
#x^2+y^2-10x-16y+71=0#