Find the equation of the circle which touches the y-axis and whose centre is (-2, -3).
Understand the Problem
The question is asking us to find the equation of a circle that touches the y-axis and has a specified center at the coordinates (-2, -3). This involves using the properties of circles, specifically relating to their radius and position in the Cartesian plane.
Answer
The equation of the circle is $(x + 2)^2 + (y + 3)^2 = 4$.
Answer for screen readers
The equation of the circle is $(x + 2)^2 + (y + 3)^2 = 4$.
Steps to Solve
- Identify the center and the radius of the circle
The center of the circle is given as the coordinates $(-2, -3)$. Since the circle touches the y-axis, the distance from the center to the y-axis is equal to the radius.
- Calculate the distance from the center to the y-axis
The distance from the center $(-2, -3)$ to the y-axis (which is the line $x = 0$) is the absolute value of the x-coordinate of the center. Thus, the radius $r$ of the circle is: $$ r = |-2| = 2 $$
- Write the standard equation of the circle
The standard equation of a circle is given by the formula: $$ (x - h)^2 + (y - k)^2 = r^2 $$ where $(h, k)$ is the center and $r$ is the radius. Substituting in the values we have: $$ (x - (-2))^2 + (y - (-3))^2 = 2^2 $$
- Simplify the equation
Now we can simplify the equation: $$ (x + 2)^2 + (y + 3)^2 = 4 $$
The equation of the circle is $(x + 2)^2 + (y + 3)^2 = 4$.
More Information
This equation represents a circle with center $(-2, -3)$ and a radius of $2$. The fact that it touches the y-axis implies that its nearest point to the y-axis is exactly $2$ units away.
Tips
- Mixing up the center coordinates and the radius.
- Forgetting that the circle just touches the y-axis, leading to the incorrect calculation of the radius.
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