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coordinate geometry

thorpelizts

New member
Sep 7, 2012
6
Find the equation of the circle whose centre is (-3,4) and which touches the y-axis.

how do i even begin?
 

Wilmer

In Memoriam
Mar 19, 2012
376
Find the equation of the circle whose centre is (-3,4) and which touches the y-axis.
how do i even begin?
Teacher gave no instructions, no teaching?
Google "equation of circle".
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Find the equation of the circle whose centre is (-3,4) and which touches the y-axis.

how do i even begin?
Hi thorpelizts, :)

Let \(P\equiv (-3,4)\) and let \(Q\) be the point of intersection of the circle and the y-axis. Since the y-axis is a tangent to the circle, \(PQ\) is perpendicular to the y-axis. Now I am sure you can find the length of \(PQ\) which is the radius of the circle. Can you give it a try?

Kind Regards,
Sudharaka.
 

soroban

Well-known member
Feb 2, 2012
409
Hello, thorpelizts!

Find the equation of the circle whose centre is (-3,4) and which touches the y-axis.

How do i even begin? . Make a sketch!

You are expected to know this formula: .[tex](x-h)^2 + (y-k)^2 \:=\:r^2[/tex]
. . where [tex](h,k)[/tex] is the center and [tex]r[/tex] is the radius.

Code:
                            |
                * * *       |
            *           *   |
          *               * |
         *                 *|
                            |
        *              r    *
        *         * - - - - *4
        *      (-3,4)       *
                            |
         *                 *|
          *               * |
            *           *   |
                * * *       |
                            |
    - - - - - - - + - - - - + - - -
                 -3         |
You know [tex]h = -3,\;k=4.[/tex]

Can you guess what the radius is?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
While soroban's method is easiest, you might also consider we want the solution of the systerm:

(x + 3)2 + (y - 4)2 = r2

x = 0

to have one real root.

Substitute into the first equation from the second:

(0 + 3)2 + (y - 4)2 = r2

(y - 4)2 + 9 - r2 = 0

We want this quadratic to have one root, hence the discriminant must be zero:

02 - 4(1)(9 - r2) = 0

r = 3
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Find the equation of the circle whose centre is (-3,4) and which touches the y-axis.

how do i even begin?
You begin by drawing a picture.

CB