Question
The equation of a circle is given by 2x^2 + 2y^2 - 8x + 5y - 10 = 0.
Find the
Coordinate of the center
Radius of the circle
Coordinate of p and q, if the circle cuts the x_axis at the points p and q
Find the
Coordinate of the center
Radius of the circle
Coordinate of p and q, if the circle cuts the x_axis at the points p and q
Answers
GPT-4o mini
To find the center, radius, and the points where the circle intersects the x-axis, we first need to rewrite the given equation of the circle in standard form.
The equation given is:
\[
2x^2 + 2y^2 - 8x + 5y - 10 = 0.
\]
### Step 1: Rewrite the equation
Divide the whole equation by 2 to simplify it:
\[
x^2 + y^2 - 4x + \frac{5}{2} y - 5 = 0.
\]
Reorganizing gives:
\[
x^2 - 4x + y^2 + \frac{5}{2}y - 5 = 0.
\]
### Step 2: Complete the square for \(x\) and \(y\)
1. **For \(x\)**:
- Take \(-4\):
- \((x^2 - 4x)\) can be completed to \((x-2)^2 - 4\).
2. **For \(y\)**:
- Take \(\frac{5}{2}\):
- \((y^2 + \frac{5}{2} y)\) can be completed to \((y + \frac{5}{4})^2 - \frac{25}{16}\).
Now substitute completed squares back into the equation:
\[
(x-2)^2 - 4 + (y + \frac{5}{4})^2 - \frac{25}{16} - 5 = 0.
\]
Simplify:
\[
(x-2)^2 + (y + \frac{5}{4})^2 - 4 - \frac{25}{16} - 5 = 0.
\]
To combine constants, convert \(-4\) and \(-5\) into sixteenths:
\[
-4 = -\frac{64}{16}, \quad -5 = -\frac{80}{16}.
\]
Thus, we have:
\[
-\frac{64}{16} - \frac{25}{16} - \frac{80}{16} = -\frac{169}{16}.
\]
Setting the equation:
\[
(x-2)^2 + (y + \frac{5}{4})^2 = \frac{169}{16}.
\]
### Step 3: Identify the center and radius
From the standard form \((x-h)^2 + (y-k)^2 = r^2\), we identify:
- Center \((h, k) = (2, -\frac{5}{4})\).
- Radius \(r = \sqrt{\frac{169}{16}} = \frac{13}{4}\).
### Step 4: Find points \(p\) and \(q\) where the circle intersects the x-axis
To find the intersection points with the x-axis, set \(y = 0\) in the equation:
\[
(x-2)^2 + (0 + \frac{5}{4})^2 = \frac{169}{16}.
\]
Substituting for \(y = 0\):
\[
(x-2)^2 + \left(\frac{5}{4}\right)^2 = \frac{169}{16}.
\]
Calculating \(\left(\frac{5}{4}\right)^2 = \frac{25}{16}\):
\[
(x-2)^2 + \frac{25}{16} = \frac{169}{16}.
\]
Subtract \(\frac{25}{16}\) from both sides:
\[
(x-2)^2 = \frac{169}{16} - \frac{25}{16} = \frac{144}{16} = 9.
\]
Taking the square root gives:
\[
x - 2 = 3 \quad \text{or} \quad x - 2 = -3.
\]
Thus,
\[
x = 5 \quad \text{or} \quad x = -1.
\]
### Conclusion
The points where the circle intersects the x-axis are:
- **Coordinates of the center**: \((2, -\frac{5}{4})\)
- **Radius**: \(\frac{13}{4}\)
- **Points \(p\) and \(q\)**: \((-1, 0)\) and \((5, 0)\) respectively.
The equation given is:
\[
2x^2 + 2y^2 - 8x + 5y - 10 = 0.
\]
### Step 1: Rewrite the equation
Divide the whole equation by 2 to simplify it:
\[
x^2 + y^2 - 4x + \frac{5}{2} y - 5 = 0.
\]
Reorganizing gives:
\[
x^2 - 4x + y^2 + \frac{5}{2}y - 5 = 0.
\]
### Step 2: Complete the square for \(x\) and \(y\)
1. **For \(x\)**:
- Take \(-4\):
- \((x^2 - 4x)\) can be completed to \((x-2)^2 - 4\).
2. **For \(y\)**:
- Take \(\frac{5}{2}\):
- \((y^2 + \frac{5}{2} y)\) can be completed to \((y + \frac{5}{4})^2 - \frac{25}{16}\).
Now substitute completed squares back into the equation:
\[
(x-2)^2 - 4 + (y + \frac{5}{4})^2 - \frac{25}{16} - 5 = 0.
\]
Simplify:
\[
(x-2)^2 + (y + \frac{5}{4})^2 - 4 - \frac{25}{16} - 5 = 0.
\]
To combine constants, convert \(-4\) and \(-5\) into sixteenths:
\[
-4 = -\frac{64}{16}, \quad -5 = -\frac{80}{16}.
\]
Thus, we have:
\[
-\frac{64}{16} - \frac{25}{16} - \frac{80}{16} = -\frac{169}{16}.
\]
Setting the equation:
\[
(x-2)^2 + (y + \frac{5}{4})^2 = \frac{169}{16}.
\]
### Step 3: Identify the center and radius
From the standard form \((x-h)^2 + (y-k)^2 = r^2\), we identify:
- Center \((h, k) = (2, -\frac{5}{4})\).
- Radius \(r = \sqrt{\frac{169}{16}} = \frac{13}{4}\).
### Step 4: Find points \(p\) and \(q\) where the circle intersects the x-axis
To find the intersection points with the x-axis, set \(y = 0\) in the equation:
\[
(x-2)^2 + (0 + \frac{5}{4})^2 = \frac{169}{16}.
\]
Substituting for \(y = 0\):
\[
(x-2)^2 + \left(\frac{5}{4}\right)^2 = \frac{169}{16}.
\]
Calculating \(\left(\frac{5}{4}\right)^2 = \frac{25}{16}\):
\[
(x-2)^2 + \frac{25}{16} = \frac{169}{16}.
\]
Subtract \(\frac{25}{16}\) from both sides:
\[
(x-2)^2 = \frac{169}{16} - \frac{25}{16} = \frac{144}{16} = 9.
\]
Taking the square root gives:
\[
x - 2 = 3 \quad \text{or} \quad x - 2 = -3.
\]
Thus,
\[
x = 5 \quad \text{or} \quad x = -1.
\]
### Conclusion
The points where the circle intersects the x-axis are:
- **Coordinates of the center**: \((2, -\frac{5}{4})\)
- **Radius**: \(\frac{13}{4}\)
- **Points \(p\) and \(q\)**: \((-1, 0)\) and \((5, 0)\) respectively.