Finding the Center and Radius - Math

Card 1 of 8

Didn't Know

Knew It

1 of 87 left

Question

Find the center and radius of the circle defined by the equation:

Tap to reveal answer

Answer

The equation of a circle is: $(x-x_${1})^2$+(y-y_${1})^2$$=r^2$$ where is the radius and is the center.

In this problem, the equation is already in the format required to determine center and radius. To find the -coordinate of the center, we must find the value of that makes equal to 0, which is 3. We do the same to find the y-coordinate of the center and find that . To find the radius we take the square root of the constant on the right side of the equation which is 6.

← Didn't Know|Knew It →

Question

Find the center and radius of the circle defined by the equation:

Tap to reveal answer

Answer

The equation of a circle is: $(x-x_${1})^2$+(y-y_${1})^2$$=r^2$$ where is the radius and is the center.

In this problem, the equation is already in the format required to determine center and radius. To find the -coordinate of the center, we must find the value of that makes equal to , which is . We do the same to find the y-coordinate of the center and find that . To find the radius we take the square root of the constant on the right side of the equation which is 10.

← Didn't Know|Knew It →

Question

Find the center and radius of the circle defined by the equation:

Tap to reveal answer

Answer

The equation of a circle is: $(x-x_${1})^2$+(y-y_${1})^2$$=r^2$$ where is the radius and is the center.

In this problem, the equation is already in the format required to determine center and radius. To find the -coordinate of the center, we must find the value of that makes equal to 0, which is 3. We do the same to find the y-coordinate of the center and find that . To find the radius we take the square root of the constant on the right side of the equation which is 6.

← Didn't Know|Knew It →

Question

Find the center and radius of the circle defined by the equation:

Tap to reveal answer

Answer

The equation of a circle is: $(x-x_${1})^2$+(y-y_${1})^2$$=r^2$$ where is the radius and is the center.

In this problem, the equation is already in the format required to determine center and radius. To find the -coordinate of the center, we must find the value of that makes equal to , which is . We do the same to find the y-coordinate of the center and find that . To find the radius we take the square root of the constant on the right side of the equation which is 10.

← Didn't Know|Knew It →

Question

Find the center and radius of the circle defined by the equation:

Tap to reveal answer

Answer

The equation of a circle is: $(x-x_${1})^2$+(y-y_${1})^2$$=r^2$$ where is the radius and is the center.

In this problem, the equation is already in the format required to determine center and radius. To find the -coordinate of the center, we must find the value of that makes equal to 0, which is 3. We do the same to find the y-coordinate of the center and find that . To find the radius we take the square root of the constant on the right side of the equation which is 6.

← Didn't Know|Knew It →

Question

Find the center and radius of the circle defined by the equation:

Tap to reveal answer

Answer

The equation of a circle is: $(x-x_${1})^2$+(y-y_${1})^2$$=r^2$$ where is the radius and is the center.

In this problem, the equation is already in the format required to determine center and radius. To find the -coordinate of the center, we must find the value of that makes equal to , which is . We do the same to find the y-coordinate of the center and find that . To find the radius we take the square root of the constant on the right side of the equation which is 10.

← Didn't Know|Knew It →

Question

Find the center and radius of the circle defined by the equation:

Tap to reveal answer

Answer

The equation of a circle is: $(x-x_${1})^2$+(y-y_${1})^2$$=r^2$$ where is the radius and is the center.

In this problem, the equation is already in the format required to determine center and radius. To find the -coordinate of the center, we must find the value of that makes equal to 0, which is 3. We do the same to find the y-coordinate of the center and find that . To find the radius we take the square root of the constant on the right side of the equation which is 6.

← Didn't Know|Knew It →

Question

Find the center and radius of the circle defined by the equation:

Tap to reveal answer

Answer

The equation of a circle is: $(x-x_${1})^2$+(y-y_${1})^2$$=r^2$$ where is the radius and is the center.

In this problem, the equation is already in the format required to determine center and radius. To find the -coordinate of the center, we must find the value of that makes equal to , which is . We do the same to find the y-coordinate of the center and find that . To find the radius we take the square root of the constant on the right side of the equation which is 10.

← Didn't Know|Knew It →

Knew It