Thread: Ahead but behind.

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

Originally Posted by Cikadamate

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

All circles have equation , where is the coordinate of the centre, and is the radius.

Do you think you can substitute all the appropriate information into this equation?

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

Well, the equation of a circle and the distance formula are in fact equivalent. Both come from Pythagoras' Theorem.

Originally Posted by Cikadamate

Thanks a lot. That was easy
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

That, applied to this problem, gives .

Squaring both sides, so that . From that, . There are two solutions because there are two such points.

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.)

I didn't know how to type subscript