Find an equation of the circle with centre C(5, -3) that passes through the point A(-2, 1) in the form (x-a)^2 + (y-b)^2 = k

step 1 remember than the a and b terms locate the centre of the circle on the axis so we can substitute in the centre values for a and b. (x-5)^2 + (y-(-3))^2 = k. (x-5)^2 + (y+3)^2 = k.

Step 2. k is a constnat representing the radius squared. calculate the radius of the circle using pythaogras. distance from centre to point A in the x direction is 5-(-2)=7. distance from centre to point B in the y direction is 1-(-3)= 4. using pythagoras we know that A^2=B^2 + C^2. this means the radius^2 = X distance^2 + Y distance^2.
so r^2 = 7^2 + 4^2. r^2 = 49+16=65.

Step 3. putting both centre component and radius together we obtain (x-5)^2 + (y+3)^2 = 65. This is the equation of the circle.

TW
Answered by Tim W. Maths tutor

5251 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

If f(x) = (3x-2) / x-5 x>6, find a.) ff(8) b.) the range of f(x) c.) f^-1(x) and state its range.


Find the value of x in (4^5⋅x+32^2)⋅2^5=2^16⋅x


Given y = 2x(x2 – 1)5, show that (a) dy/dx = g(x)(x2 – 1)4 where g(x) is a function to be determined. (b) Hence find the set of values of x for which dy/dx > 0


The quadratic equation 2x^2 + 8x + 1 = 0 has roots a and b. Write down the value of a + b, a*b and a^2 + b^2.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning