How do you find the general form of the equation of this circle given Center (2,3) and tangent to the x-axis?

2 Answers
Mark D.
May 1, 2018

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

Explanation:

If the centre is at (2,3) and the x axis is a tangent to the circle then the radius must be 3.

#(x-a)^2 + (y-b)^2 = r^2# is the general formula for a circle with centre #(a,b)# and radius #r#

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

Jim G.
May 1, 2018

#x^2+y^2-4x-6y+4=0#

Explanation:

#"the general form of the equation of a circle is"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(x^2+y^2+2gx+2fy+c=0)color(white)(2/2)|)))#

#"with centre "=(-g,-f)" and"#

#"and radius "=sqrt(g^2+f^2-c)color(white)(x);g^2+f^2-c>0#

#"to begin obtain the equation in standard form"#

#•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)#

#"and "r=3to"(distance from x-axis to y-coord of centre)"#

#rArr(x-2)^2+(y-3)^2=9larrcolor(blue)"in standard form"#

#"distributing and rearranging"#

#x^2-4x+4+y^2-6y+9-9=0#

#rArrx^2+y^2-4x-6y+4=0larrcolor(red)"in general form"#

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