How do you write the equation for a circle with center at (2,3) that is tangent to the x-axis?

2 Answers

Answer:

#(x-2)^2+(y-3)^2=9#

Explanation:

Given that the center of circle is #(x_1, y_1)\equiv(2, 3)#.

The above circle is tangent to the x-axis hence its radius

#r=\text{y-coordinate}=3#

hence the equation of the circle is given by following formula

#(x-x_1)^2+(y-y_1)^2=r^2#

#(x-2)^2+(y-3)^2=3^2#

#(x-2)^2+(y-3)^2=9#

Jul 26, 2018

Answer:

#(x-2)^2+(y-3)^2=9#

Explanation:

#"the equation of a circle in standard form is"#

#•color(white)(x)(x-a)^2+(y-b)^2=r^2#

#"where "(a,b)" are the coordinates of the centre and r"#
#"is the radius"#

#"here "(a,b)=(2,3)#

#"r is the vertical distance from the x-axis to the centre"#

#r="y-coordinate of centre "=3#

#(x-2)^2+(y-3)^2=9larrcolor(red)"equation of circle"#