Find the radius of a circle with center (4,1) if a chord of length 4 sqrt of 2 is bisected at (7,4)?

Question

Find the radius of a circle with center (4,1) if a chord of length 4 sqrt of 2 is bisected at (7,4)?

Answer 1

start by drawing a line between the chord's midpoint and the center of the circle. the length of that line (also known as the apothem fyi) is the distance between (4,1) and (7,4). the distance formula for this yields 3*sqrt(2).

the distance between the midpoint of the chord and either point where it intersects the circle is 2*sqrt(2) (half the length of the chord).

now what we have here is a right triangle with one leg = 2*sqrt(2) and the other leg = 3*sqrt(2). the hypotenuse connects the center of the circle with a point of intersection between the chord and circle, and is therefore the radius we're looking for.

r^2 = (3*sqrt(2))^2 + (2*sqrt(2))^2
r = sqrt(26).

(it's much easier to see with a graph)

Answer 2

Draw 2 radii of fifty 8 gadgets and connect the standards the place they intersect the circumference of the circle. to variety a chord. Draw the perpendicular from the middle of the circle to the chord. be conscious you have 2 congruent proper triangles, each with hypotenuse fifty 8 and height 40. call the the rest leg of each triangle x so as that the dimensions of the chord is 2x. on condition that x is the leg of a suitable triangle with hypotenuse fifty 8 and different leg of 40 via Pythagoras x = sqrt(fifty 8^2 - 40^2) = 40 two. The chord is 2x = 2(40 two) = eighty 4 gadgets

Answer 3

r^2 = 3^2 + 3^2 + 8 = 9 + 9 + 8 = 26
r = √26