Question

Find the center \((h, k)\) and radius \(r\) of each circle; \((b)\) graph each circle; \((c)\) find the intercepts, if any. $$ 3(x+1)^{2}+3(y-1)^{2}=6 $$

Short answer

Center: (-1, 1), Radius: \(\sqrt{2}\), X-Intercepts: 0, -2, Y-Intercepts: 2, 0.

Step-by-step solution

Simplify the Equation

Start by simplifying the given equation. Divide both sides by 3 to simplify. \[ \frac{3(x+1)^{2} + 3(y-1)^{2}}{3} = \frac{6}{3} \] This reduces to: \[ (x+1)^{2} + (y-1)^{2} = 2 \]

Identify the Center and Radius

Rewrite the simplified equation in the standard form of a circle equation \[ (x-h)^{2} + (y-k)^{2} = r^{2} \]. From \[ (x+1)^{2} + (y-1)^{2} = 2 \], we can see that \[ h = -1 \], \[ k = 1 \], and \[ r^{2} = 2 \]. Therefore, \[ r = \sqrt{2} \]. So, the center \(h, k\) is \((-1, 1)\) and the radius \(r\) is \(\sqrt{2}\).

Graph the Circle

To graph the circle, plot the center at \((-1,1)\) on a coordinate plane. Then use the radius \(\sqrt{2}\) to draw points around the center and draw the circle.

Find the X-Intercepts

To find the x-intercepts, set \(y=0\) in the simplified equation: \[ (x+1)^{2} + (0-1)^{2} = 2 \]. This simplifies to: \[ (x+1)^{2} + 1 = 2 \] , and then \[ (x+1)^{2} = 1 \]. Solving for \(x\), we get: \[ x+1 = \pm 1 \], \[ x = 0 \] or \[ x = -2 \]. Therefore, the x-intercepts are \( x = 0 \) and \(x =-2)\).

Find the Y-Intercepts

To find the y-intercepts, set \(x=0\) in the equation: \[ (0+1)^{2} + (y-1)^{2} = 2 \]. This simplifies to: \[ 1 + (y-1)^{2} = 2 \] , and then \[ (y-1)^{2} = 1 \]. Solving for \(y\), we get: \[ y-1 = \pm 1 \], \[ y = 2 \] or \[ y = 0 \]. Therefore, the y-intercepts are \( y = 2 \) and \(y = 0\).

Key concepts

center of circle

To determine the center of a circle from its equation, you need to rewrite the equation in the *standard form* of a circle.
This is \((x - h)^{2} + (y - k)^{2} = r^{2}\). The terms \(h\) and \(k\) represent the center of the circle \((h, k)\).

By transforming the given equation \((x+1)^{2} + (y-1)^{2} = 2\) into standard form, it is clear that \((x+1)\) and \((y-1)\) correspond to \h = -1\ and \k = 1\, respectively.

Thus, the center of the circle is \((-1, 1)\). The transformation involves recognizing the shift from the origin, giving the values of \(h\) and \(k\). This method makes it easier to identify the exact location of the circle on a coordinate plane.

radius calculation

Calculating the radius of a circle from its standard form equation \((x - h)^{2} + (y - k)^{2} = r^{2}\) requires identifying the value of \(r^{2}\).

In the given simplified equation \((x+1)^{2} + (y-1)^{2} = 2\), we see that \r^{2} = 2\.
To find \(r\), take the square root:

\r = \sqrt{2}\.

The radius \(r\) thus has a length of \sqrt{2}\ units. Understanding this is crucial for both graphing the circle and performing further calculations, such as finding intercepts or evaluating other points on the circle. By identifying \(r^{2}\) and taking its square root, you transform the equation into practical terms, simplifying the graphing and interpretation process.

intercepts

Intercepts are the points where the circle crosses the x-axis and y-axis.

Finding X-Intercepts:
To find the x-intercepts, set \(y\) to 0 in the equation:
\((x+1)^{2} + (0-1)^{2} = 2\)

The equation simplifies to:
\((x+1)^{2} + 1 = 2\)
Then:
\((x+1)^{2} = 1\)
Solve for \(x\) to get:
\x + 1 = \pm 1\
Therefore, the possible values of \(x\) are 0 and -2.

The x-intercepts are thus at \(x = 0\) and \(x = -2\).

Finding Y-Intercepts:
To find the y-intercepts, set \(x\) to 0 in the equation:
\((0+1)^{2} + (y-1)^{2} = 2\)
The equation simplifies to:
\(1 + (y-1)^{2} = 2\)
Then:
\(y-1)^{2} = 1\)
Solve for \(y\) to get:
\y - 1 = \pm 1\
Therefore, the possible values of \(y\) are 2 and 0.

The y-intercepts are thus at \(y = 2\) and \(y = 0\).
Intercepts are important in visualizing where a circle intersects the coordinate axes, helping give a clearer picture of its orientation and size.