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

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SUMMARY

The equation of the circle centered at (-3, 4) that touches the y-axis is derived using the standard circle equation \((x-h)^2 + (y-k)^2 = r^2\). Given that the center is (-3, 4), the radius \(r\) is determined to be 3, as the distance from the center to the y-axis is 3 units. Thus, the equation of the circle is \((x + 3)^2 + (y - 4)^2 = 9\). This conclusion is reached by ensuring the quadratic formed by substituting the y-axis into the circle's equation has a discriminant of zero, confirming tangency.

PREREQUISITES
  • Understanding of the standard form of a circle's equation
  • Knowledge of the concept of tangents in geometry
  • Ability to calculate the discriminant of a quadratic equation
  • Familiarity with coordinate geometry
NEXT STEPS
  • Study the derivation of the standard circle equation in coordinate geometry
  • Learn about the properties of tangents and their relationship with circles
  • Explore quadratic equations and the significance of the discriminant
  • Practice solving problems involving circles and tangents in various coordinate systems
USEFUL FOR

Students learning geometry, mathematics educators, and anyone interested in mastering the concepts of circles and tangents in coordinate geometry.

thorpelizts
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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
 

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