MHB Equation of Circle Centered at (-3,4) Touches Y-Axis

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To find the equation of a circle centered at (-3, 4) that touches the y-axis, the radius must equal the distance from the center to the y-axis, which is 3. The standard form of the circle's equation is (x + 3)² + (y - 4)² = r². Substituting the radius into the equation gives (x + 3)² + (y - 4)² = 9. A visual sketch is recommended to aid understanding of the circle's position relative to the y-axis.
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Find the equation of the circle whose centre is (-3,4) and which touches the y-axis.

how do i even begin?
 
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thorpelizts said:
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".
 
thorpelizts said:
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.
 
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: .(x-h)^2 + (y-k)^2 \:=\:r^2
. . where (h,k) is the center and r is the radius.

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

Can you guess what the radius is?
 
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
 
thorpelizts said:
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
 

Thread 'Trigonometry problem of interest'

Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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