Asked by Angelica Ochoa on May 19, 2024
Verified
Write the standard form of the equation of the circle with the center at (0,2) that passes through the point (7,3) .
A) x2+(y+2) 2=50x ^ { 2 } + ( y + 2 ) ^ { 2 } = 50x2+(y+2) 2=50
B) x2+(y−2) 2=58x ^ { 2 } + ( y - 2 ) ^ { 2 } = 58x2+(y−2) 2=58
C) x2+(y−2) 2=50x ^ { 2 } + ( y - 2 ) ^ { 2 } = 50x2+(y−2) 2=50
D) (x−2) 2+y2=58( x - 2 ) ^ { 2 } + y ^ { 2 } = 58(x−2) 2+y2=58
E) (x+2) 2+y2=50( x + 2 ) ^ { 2 } + y ^ { 2 } = 50(x+2) 2+y2=50
Center
A point of origin or midpoint around which a figure or object is symmetrically arranged or balanced.
Point
A precise location or position in a geometric space, not having any dimension.
- Attain proficiency in constructing and understanding the standard expressions of circle equations.
Verified Answer
AD
Angelina DiColaMay 21, 2024
Final Answer :
C
Explanation :
The standard form of the equation of a circle with center $(h,k)$ and radius $r$ is:
(x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2(x−h)2+(y−k)2=r2
In this case, the center is $(0,2)$, so $h=0$ and $k=2$. We need to find the radius, which is the distance between the center $(0,2)$ and the point $(7,3)$:
r=(7−0)2+(3−2)2=50r=\sqrt{(7-0)^2+(3-2)^2}=\sqrt{50}r=(7−0)2+(3−2)2=50
So, the equation of the circle is:
(x−0)2+(y−2)2=50(x-0)^2+(y-2)^2=50(x−0)2+(y−2)2=50
Simplifying, we get:
x2+(y−2)2=50x^2+(y-2)^2=50x2+(y−2)2=50
which matches choice C.
(x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2(x−h)2+(y−k)2=r2
In this case, the center is $(0,2)$, so $h=0$ and $k=2$. We need to find the radius, which is the distance between the center $(0,2)$ and the point $(7,3)$:
r=(7−0)2+(3−2)2=50r=\sqrt{(7-0)^2+(3-2)^2}=\sqrt{50}r=(7−0)2+(3−2)2=50
So, the equation of the circle is:
(x−0)2+(y−2)2=50(x-0)^2+(y-2)^2=50(x−0)2+(y−2)2=50
Simplifying, we get:
x2+(y−2)2=50x^2+(y-2)^2=50x2+(y−2)2=50
which matches choice C.
Learning Objectives
- Attain proficiency in constructing and understanding the standard expressions of circle equations.