# The center of a circle is at (-3,-2) and its radius is 7. Find the length of the chord which is bisected at (3,1). At what points does the circle cut the y - axis?

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### 1 Answer

Given a circle centered at (-3,-2) with radius 7:

(1) Let the center of the circle be C. Let `bar(AB)` be a chord of the circle such that its midpoint M is at (3,1). Find `AB` :

Recall that a radius drawn to the midpoint of a chord is perpendicular to the chord. (Two points that are equidistant from the endpoints of a segment lie on the perpendicular bisector of the segment.)

We find CM using the distance formula: `CM=sqrt((-3-3)^2+(-2-1)^2)=sqrt(36+9)=sqrt(45)=3sqrt(5)`

Note that CA=7 (it is a radius) and that `Delta CMA` is a right triangle. Using the Pythagorean theorem we find AM:

`AM^2+CM^2=CA^2`

`AM^2+(3sqrt(5))^2=7^2`

`AM^2+45=49`

`AM=2`

Since M is the midpoint of `bar(AB)` , we have AB=4.

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**The length of the chord bisected at (3,1) of the given circle is 4.**

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(2) To find the y-intercepts of the circle, we write its equation in standard form:

`(x+3)^2+(y+2)^2=49`

The y-intercepts occur when x=0, so:

`9+(y+2)^2=49`

`y^2+4y-36=0`

`y=(-4+-sqrt(16-4(1)(-36)))/2`

`y=(-4+-sqrt(160))/2`

`y=-2+-2sqrt(10)~~-2+-6.32`

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**The y-intercepts of the circle are at `-2+2sqrt(10),-2-2sqrt(10)` or approximately `4.32,-8.32` **

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