• Jun 11th 2012, 03:13 AM
I'm Cikadamate, I joined just because I am a year ahead in maths but I'm struggling. I am hoping that when necessary I will be able to use this forum to help me understand anything I am stuck on.
Right now we're doing co-ordinate geometry and This is the question which prompted me to join:

A circle has centre G(2,-1). A point H(a,3) on the circumference of the circle is joined to the centre of the circle. if the radius of the circle is 8, find 'a'.

What I need to know is the method for working this type of question out.
Any help would be most appreciated :D
• Jun 11th 2012, 03:17 AM
Prove It
Quote:

I'm Cikadamate, I joined just because I am a year ahead in maths but I'm struggling. I am hoping that when necessary I will be able to use this forum to help me understand anything I am stuck on.
Right now we're doing co-ordinate geometry and This is the question which prompted me to join:

A circle has centre G(2,-1). A point H(a,3) on the circumference of the circle is joined to the centre of the circle. if the radius of the circle is 8, find 'a'.

What I need to know is the method for working this type of question out.
Any help would be most appreciated :D

All circles have equation \displaystyle \begin{align*} (x - h)^2 + (y - k)^2 = r^2 \end{align*}, where \displaystyle \begin{align*} (h,k) \end{align*} is the coordinate of the centre, and \displaystyle \begin{align*} r \end{align*} is the radius.

Do you think you can substitute all the appropriate information into this equation?
• Jun 11th 2012, 03:28 AM
Thanks a lot. That was easy :D
I was trying to solve it using the distance formula d = √(x-x)²+(y-y)²
I confused me.
• Jun 11th 2012, 03:42 AM
Prove It
Well, the equation of a circle and the distance formula are in fact equivalent. Both come from Pythagoras' Theorem.
• Jun 11th 2012, 07:54 AM
HallsofIvy
Quote:

Thanks a lot. That was easy :D
I was trying to solve it using the distance formula d = √(x-x)²+(y-y)²
I confused me.

Well, that would confuse me because it is using the same letters, x and y, to represent two different points.

Better is $d= \sqrt{(x_1- x_0)^2+ (y_1- y_0)^2}$

That, applied to this problem, gives $8= \sqrt{(a- 2)^2+ (3-(-1))^2+}= \sqrt{(a- 2)^2+ 4^2}$.

Squaring both sides, $64= (a- 2)^2+ 16$ so that $(a- 2)^2= 64- 16= 48$. From that, $a- 2= \pm\sqrt{48}= \pm4\sqrt{3}$. There are two solutions because there are two such points.
• Jun 11th 2012, 05:35 PM
Deveno
i just drew a circle centered at (2,-1) of radius 8. we want to know the x-coordinate(s) of the point(s) where this circle intersects the line y = 3.

if one draws a right triangle, formed by (2,-1), (a,-1) and (a,3), one sees immediately that the hypotenuse has length 8, the vertical leg has height 4 ( = 3 - (-1)), and the horizontal leg has length |a-2| (a could be to the left of the center).

by the pythagorean theorem:

(a-2)2 + 16 = 64, just as HallsofIvy derived.

(EDIT: there are good reasons why all three answers given to you will lead to the same answer. in a sense, circles ARE "distance measuring curves", that is: the formula for distance, and the formula for the equation of a circle are two different ways to interpret the pythagorean equation:

A2 + B2 = C2.

circles hold "C" constant, and let "A" and "B" vary, while distance interprets "C" as a function of "A" and "B". in other words, circles implicitly define one leg length of the inscribed right triangles as a function of the other leg's length, while the distance formula explicitly calculates the hypotenuse, given the legs.

i often find it amazing that "pointy" things like triangles, and "line length" things (like distance) are so intimately bound up with the notion of a circle, which is neither pointy nor "linear". this "pythagorean connection" often shows up in unexpected areas in many branches of mathematics. it's THAT deep.)
• Jun 12th 2012, 03:37 AM