Write down the coordinates of the centre and the radius of the circle with equation x^2+y^2-4x-8y+11=0
The general equation for a circle is as follows:(x-a)^2+(y-b)^2=r^2where (a,b) are the coordinates of the centre of the circle and r is the radius of the circle.x^2+y^2-4x-8y+11=0In this case, group all x terms and y terms together, giving: x^2-4x+y^2-8y+11=0Then, complete the square on the x terms and then on the y terms:(x-2)^2-4+(y-4)^2-16+11=0Add all the numbers outside the brackets together:(x-2)^2+(y-4)^2-9=0Move 9 to the right hand side so we have our equation in the form of the general equation:(x-2)^2+(y-4)^2=9Comparing to the general equation:a=2,b=4 so the coordinates of the centre are (2,4)r^2=9 so r=3*note that r cannot be -3 because it describes a distance and you cannot have a negative distance.
MC
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