1. ## Circles and Chords

Having a bit of trouble with this one. Can anyone help me out?

Many thanks.

Q.
C is a circle with centre (-1, -4). The midpoint of a chord of length $\displaystyle 2\sqrt{5}$ is (2, 0). Find the length of the radius of C.

Attempt: Perpendicular distance [bc] from centre c to x-axis is 4,
Let |ab| = 1/2 length of chord on x-axis = $\displaystyle \sqrt{5}$,
$\displaystyle |ac|^2=(\sqrt{5})^2+4^2=21\rightarrow |ac|=\sqrt{21}=$ radius,

Ans:
(From text book): Radius = $\displaystyle \sqrt{30}$

2. ## Re: Circles and Chords

Originally Posted by GrigOrig99
Q.[/B] C is a circle with centre (-1, -4). The midpoint of a chord of length $\displaystyle 2\sqrt{5}$ is (2, 0). Find the length of the radius of C.

Attempt: Perpendicular distance [bc] from centre c to x-axis is 4
So I assume b is (-1, 0).

Originally Posted by GrigOrig99
Let |ab| = 1/2 length of chord on x-axis = $\displaystyle \sqrt{5}$
I am not sure what the "length of chord on x-axis" is: the projection of the chord to the x-axis? Why are you interested in it since it is not known that the chord is parallel to the x-axis? Also, "Let |ab| = $\displaystyle \sqrt{5}$" is ambiguous. Does this mean that $\displaystyle a = (-1-\sqrt{5},0)$ or $\displaystyle (-1+\sqrt{5},0)$, or maybe $\displaystyle a$ does not lie on the x-axis at all?

Hint: The line from the circle center to the chord center is perpendicular to the chord. Using this fact, find a right triangle with a radius as the hypotenuse.

3. ## Re: Circles and Chords

Let |ab| = 1/2 length of chord = $\displaystyle \frac{2\sqrt{5}}{2}}=\sqrt{5}$,
With centre c = (-1, 4), calculate perpendicular distance to cord via (2, 0) = $\displaystyle \sqrt{(-1-2)^2+(-4-0)^2}=\sqrt{25}=5$,
$\displaystyle |ac|^2=\sqrt{5}^2+5^2=30\rightarrow |ac|=\sqrt{30}$
Equation of a circle: $\displaystyle (x+1)^2+(y+4)^2=\sqrt{30}^2\rightarrow x^2+y^2+2x+8y-13=0$