Which equation represents the circle described? The radius is 2 units. The center is the same as the center of a circle whose equation is x2 + y2 – 8x – 6y + 24 = 0.
The Correct Answer and Explanation is:
Correct Answer:
(x – 4)² + (y – 3)² = 4
Explanation:
We are given two important pieces of information:
- The radius of the new circle is 2 units.
- The center is the same as that of another circle whose equation is:
x2+y2−8x−6y+24=0x^2 + y^2 – 8x – 6y + 24 = 0x2+y2−8x−6y+24=0
Step 1: Find the center of the given circle
We start by rewriting the equation in standard form by completing the square:
Given:
x2+y2−8x−6y+24=0x^2 + y^2 – 8x – 6y + 24 = 0x2+y2−8x−6y+24=0
Group x and y terms:
(x2−8x)+(y2−6y)=−24(x^2 – 8x) + (y^2 – 6y) = -24(x2−8x)+(y2−6y)=−24
Complete the square:
- For x2−8xx^2 – 8xx2−8x, take half of 8 (which is 4), square it: 42=164^2 = 1642=16
- For y2−6yy^2 – 6yy2−6y, take half of 6 (which is 3), square it: 32=93^2 = 932=9
Add 16 and 9 to both sides:
(x2−8x+16)+(y2−6y+9)=−24+16+9(x^2 – 8x + 16) + (y^2 – 6y + 9) = -24 + 16 + 9(x2−8x+16)+(y2−6y+9)=−24+16+9
Simplify:
(x−4)2+(y−3)2=1(x – 4)^2 + (y – 3)^2 = 1(x−4)2+(y−3)2=1
So, the center is at (4, 3) and the radius is 1 (from the original circle).
Step 2: Write the new circle’s equation
The new circle has the same center (4, 3) but a radius of 2 units.
Using the standard form of a circle:
(x−h)2+(y−k)2=r2(x – h)^2 + (y – k)^2 = r^2(x−h)2+(y−k)2=r2
Plug in the values:
(x−4)2+(y−3)2=22=4(x – 4)^2 + (y – 3)^2 = 2^2 = 4(x−4)2+(y−3)2=22=4
Final Answer:
(x – 4)² + (y – 3)² = 4
This equation represents a circle centered at (4, 3) with a radius of 2 units.
