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Question: How do you find the standard equation of the circle with endpoints of a diameter \((3,6)\) and \(\,(...

How do you find the standard equation of the circle with endpoints of a diameter (3,6)(3,6) and  (−1,4)\,( - 1,4) ?

Explanation

Solution

Hint : First we will evaluate the centre of the circle using the standard equation of circle (x−a)2+(y−b)2=r2{(x - a)^2} + {(y - b)^2} = {r^2}, where (a,b)(a,b) is the centre of the circle and rr is the radius of the circle.
Use the midpoint formula to evaluate the centre of the circle and hence the radius of the circle.

Complete step-by-step answer :
The general equation of circle is given by (x−a)2+(y−b)2=r2{(x - a)^2} + {(y - b)^2} = {r^2}
Where, (a,b)(a,b) is the centre of the circle and rr is the radius of the circle.
First, we can find the centre point of the circle using the midpoint formula as we have the values of the endpoints of the diameter.
The midpoint formula is given by:
M=(x1+x22,y1+y22)M = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)
Where x1,x2 {x_1},{x_2}\,and y1,y2{y_1},{y_2} are the endpoints of the segment.
Now, substitute the values in the midpoint formula to evaluate the centre of the circle.

M=(x1+x22,y1+y22) M=(3+(−1)2,6+42) M=(3−12,6+42) M=(22,102) M=(1,5)   M = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right) \\\ M = \left( {\dfrac{{3 + ( - 1)}}{2},\dfrac{{6 + 4}}{2}} \right) \\\ M = \left( {\dfrac{{3 - 1}}{2},\dfrac{{6 + 4}}{2}} \right) \\\ M = \left( {\dfrac{2}{2},\dfrac{{10}}{2}} \right) \\\ M = \left( {1,5} \right) \;

Hence, the centre of the circle is (1,5)(1,5).
Now, as we know the endpoints and the centre of the circle, we can calculate the length of the diameter and hence accordingly we can evaluate the radius of the circle.
For evaluating the distance between the two endpoints we will use the distance formula.
Distance formula is given by,
d=(x2−x1)2+(y2−y1)2d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}
Now substitute the values in the distance formula and solve for the value of diameter.

d=(x2−x1)2+(y2−y1)2 d=(3−1)2+(6−5)2 d=(2)2+(1)2 d=4+1 d=5   d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \\\ d = \sqrt {{{(3 - 1)}^2} + {{(6 - 5)}^2}} \\\ d = \sqrt {{{(2)}^2} + {{(1)}^2}} \\\ d = \sqrt {4 + 1} \\\ d = \sqrt 5 \;

Hence, the radius of the circle is 5\sqrt 5 .
Now we substitute all these values in the general equation of circle which is given by (x−a)2+(y−b)2=r2{(x - a)^2} + {(y - b)^2} = {r^2}
(x−1)2+(y−5)2=(5)2 (x−1)2+(y−5)2=5   {(x - 1)^2} + {(y - 5)^2} = {(\sqrt 5 )^2} \\\ {(x - 1)^2} + {(y - 5)^2} = 5 \;
So, the correct answer is “(x−1)2+(y−5)2=5{(x - 1)^2} + {(y - 5)^2} = 5”.

Note : While solving for the centre of the circle, substitute the values in the midpoint formula along with their signs. While comparing values of terms with the general equation, compare along with their respective signs. Substitute values carefully in the formulas for evaluating values of terms.