How do you write the equation of a circle centered at (-5,2) and contains the points (7,3)?

1 Answer
Jan 8, 2017

Answer:

#(x+5)^2+(y-2)^2=145#

Explanation:

The formula of a circle centred at the origin is #x^2+y^2=r^2# because you are using the principle of Pythagoras on a triangle.

When the centre of the circle is not at the origin you mathematically move it back to centre.

So we have

#(x+5)^2+(y-2)^2=r^2#

Now we need to find the length #r# and this is achieved by calculating distance from centre to the given point using the well known
#a^2+b^2=c^2# for a triangle

#r=sqrt((-5-7)^2+(2-3)^2) = sqrt(12^2+1^2) =>r^2=145#

So we now have:

#(x+5)^2+(y-2)^2=145#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Comment:")#

#x_("centre")+5 = -5+5=0 larr" moved to origin"#
#y_("centre")-2 = +2-2=0 larr" moved to origin"#

Tony B