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Question: How do you write the equation for a circle with points \(\left( {0,2a} \right),\left( {2b,0} \right)...

How do you write the equation for a circle with points (0,2a),(2b,0)\left( {0,2a} \right),\left( {2b,0} \right) as ends of a diameter?

Explanation

Solution

Centre is the midpoint of diameter and distance from centre to either of the endpoints of diameter is radius. Thereafter, we will use the midpoint formula in the given values. Further, we will use distance formula to find the equation of the circle given below:
x=x1+x22,y=y1+y22x = \dfrac{x_1 + x_2}{2},y = \dfrac{y _1 + y_2}{2} (midpoint formula)
(x−x1)2+(y−y1)2=r2{\left( {x - {x_1}} \right)^2} + {\left( {y - {y_1}} \right)^2} = {r^2} (Centre radius form of circle)
⇒r=(x2−x1)2+(y2−y1)2\Rightarrow r = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} (Distance formula)

Complete step by step answer:
Given diametric ends of the circle are (0,2a),(2b,0)\left( {0,2a} \right),\left( {2b,0} \right).
Using the midpoint formula, we calculate the coordinate of the centre first.
x=x1+x22,y=y1+y22x = \dfrac{x_1 + x_2}{2},y = \dfrac{y _1 + y_2}{2} --(1)
Where x1=0,y1=2a,x2=2b,y2=0{x_1} = 0,{y_1} = 2a,{x_2} = 2b,{y_2} = 0
Using these values in formula (1) we have
x=0+2b2,y=2a+02x = \dfrac{{0 + 2b}}{2},y = \dfrac{{2a + 0}}{2}
⇒x=b,y=a\Rightarrow x = b,y = a
Thus, the coordinate of the centre is (b,a)\left( {b,a} \right).
We apply distance formula between centre of the circle and either end of diameter to calculate its radius:
D=(x2−x1)2+(y2−y1)2D = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}
Here, x1=0,y1=2a,x2=b,y2=a{x_1} = 0,{y_1} = 2a,{x_2} = b,{y_2} = a
Using in above formula we have
⇒D=(b−0)2+(a−2a)2\Rightarrow D = \sqrt {{{\left( {b - 0} \right)}^2} + {{\left( {a - 2a} \right)}^2}}
⇒D=a2+b2\Rightarrow D = \sqrt {{a^2} + {b^2}}
Therefore, the radius of the circle is a2+b2\sqrt {{a^2} + {b^2}} .
Now, we use the centre radius formula to find the equation of the circle.
(x−x1)2+(y−y1)2=r2{\left( {x - {x_1}} \right)^2} + {\left( {y - {y_1}} \right)^2} = {r^2}
Where (x1,y1)\left( {{x_1},{y_1}} \right) is the coordinate of the centre and rr be the radius of the circle.
Using value of (x1,y1)=(b,a)\left( {{x_1},{y_1}} \right) = \left( {b,a} \right) and r=a2+b2r = \sqrt {{a^2} + {b^2}} in above formula.
⇒(x−b)2+(y−a)2=a2+b2\Rightarrow {\left( {x - b} \right)^2} + {\left( {y - a} \right)^2} = {a^2} + {b^2}
Using algebraic identity(a−b)2=a2−2ab+b2{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}
⇒x2−2xb+b2+y2−2ay+a2=a2+b2\Rightarrow {x^2} - 2xb + {b^2} + {y^2} - 2ay + {a^2} = {a^2} + {b^2}
⇒x2+y2−2bx−2ay=0\Rightarrow {x^2} + {y^2} - 2bx - 2ay = 0

Hence, the required equation of the circle is x2+y2−2bx−2ay=0{x^2} + {y^2} - 2bx - 2ay = 0.

Note: We can also use formula (x−x1)(x−x2)+(y−y1)(y−y2)=0\left( {x - {x_1}} \right)\left( {x - {x_2}} \right) + \left( {y - {y_1}} \right)\left( {y - {y_2}} \right) = 0 to find equation of circle.
Here, (x1,y1)\left( {{x_1},{y_1}} \right) and (x2,y2)\left( {{x_2},{y_2}} \right) are diametric ends.