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Question: How do you write the equation for a circle with center at \(\left( {2,3} \right)\) that is tangent t...

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

Explanation

Solution

In this question we are asked to find the equation of circle whose radius is given, and to so this we use the standard form of equation of circle which is given by (xh)2+(yk)2=r2{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}, where (h,k)\left( {h,k} \right) is the centre of the circle andrris the radius of the circle, and tangent to x-axis means the radius will be y-coordinate of the center i.e., 3, now substituting the values in the equation of the circle we will get the required equation.

Complete step by step answer:
Given radius is (2,3)\left( {2,3} \right),
So we know that standard form of equation of circle is given by (xh)2+(yk)2=r2{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}, where (h,k)\left( {h,k} \right) is the centre of the circle and ris the radius of the circle,
Now here (h,k)=(2,3)\left( {h,k} \right) = \left( {2,3} \right) i.e., h=2h = 2and k=3k = 3,
And also it is given that the equation is tangent to the x-axis, so we know that if the equation is tangent to x-axis that means the radius will be y-coordinate of the center i.e., 3, r=3r = 3,
Now substituting the values in the equation of circle we get,
(x2)2+(y3)2=32\Rightarrow {\left( {x - 2} \right)^2} + {\left( {y - 3} \right)^2} = {3^2},
Now simplifying we get,
(x2)2+(y3)2=9\Rightarrow {\left( {x - 2} \right)^2} + {\left( {y - 3} \right)^2} = 9,
So, the equation of the circle is (x2)2+(y3)2=9{\left( {x - 2} \right)^2} + {\left( {y - 3} \right)^2} = 9.

\therefore The equation for a circle with center at (2,3)\left( {2,3} \right) that is tangent to the x-axis is given by (x2)2+(y3)2=9{\left( {x - 2} \right)^2} + {\left( {y - 3} \right)^2} = 9.

Note: In solving these type of questions we should know the general equation of the circle (xh)2+(yk)2=r2{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2} also it is important to note that sometimes we should use the complete the square method with respect to xx and yy so that we obtain the equation in general from. And another general equation of a circle is x2+y2+2gx+2fy+c=0{x^2} + {y^2} + 2gx + 2fy + c = 0.