Locus HELP! (1 Viewer)

[A Leaf]

Ok, this question from an internal past paper is confusing me...

The point (6,7) lies on the circumference of a circle with centre (3,4). Determine the equation of this circle. (2 marks)

The part I don't understand is why is the radius bit, so the answers say 3 units which I get (since 6-3=3 lol) but why don't I get 3 when I sub the points into the distance formula? I don't know if I'm being dumb and making the same mistake over and over again or? If anyone could explain this weird situation that'd be great

 

[Sp3ctre]

Pretty sure the answer is wrong. If you want to go about the long way using distance formula, you can still work out the radius straight away

r^2 = (6-3)^2 + (7-4)^2
r^2 = 9 + 9
r^2 = 18

Therefore equation of the circle is (x-3)^2 + (y-4)^2 = 18

And the radius of the circle is simply sqrt18 or 3*sqrt2

Not sure how the answers get 3 for radius, and also you can't justify 3 using 6-3 because that's only taking the horizontal distance between the two points, which isn't the value of the radius

 

[A Leaf]

Pretty sure the answer is wrong. If you want to go about the long way using distance formula, you can still work out the radius straight away

r^2 = (6-3)^2 + (7-4)^2
r^2 = 9 + 9
r^2 = 18

Therefore equation of the circle is (x-3)^2 + (y-4)^2 = 18

And the radius of the circle is simply sqrt18 or 3*sqrt2

Not sure how the answers get 3 for radius, and also you can't justify 3 using 6-3 because that's only taking the horizontal distance between the two points, which isn't the value of the radius

IKR I was gobsmacked :/ but then again my school's past papers have had sooo many incorrect solutions and OOH yeah sorry I realise the 6-3 justification is wrong now haha