Question

How do you write the equation of the circle with endpoints of its diameter at $\left( { - 4,7} \right)$ and $\left( {8, - 9} \right)$ ?

Answer 1

Hint: Here we can use the standard form of the equation of a circle to find the equation of the circle. Circle equation must need the centre and radius. We first find the centre by using the midpoint formula and then find the radius by using the distance formula with the centre and one of the endpoints. Finally substitute the centre and radius into the standard form of the circle equation.

Formula used: The standard form of the equation of a circle ${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$ where $\left( {a,b} \right)$ are the coordinates of the centre and $r$ the radius.
Midpoint formula $\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}$ where $\left( {{x_1},{y_1}} \right)$and$\left( {{x_2},{y_2}} \right)$are two points.
Distance formula, $r = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $ where $\left( {{x_1},{y_1}} \right)$and$\left( {{x_2},{y_2}} \right)$ are two points.

Complete step-by-step solution:
The standard form of the equation of a circle ${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$ where $\left( {a,b} \right)$ are the coordinates of the centre and $r$ the radius.
To establish the equation, we require to know its centre and radius.
Since we are given the endpoints of the diameter. Then the centre will be at the midpoint and the radius will be the distance from the centre to either of the two endpoints.
The midpoint can be calculated using the midpoint formula that is a centre,
$\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}$ Where $\left( {{x_1},{y_1}} \right)$and$\left( {{x_2},{y_2}} \right)$ are two points.
The two points are $\left( { - 4,7} \right)$ and $\left( {8, - 9} \right)$
So, ${x_1} = - 4,{y_1} = 7,\,{x_2} = 8,\,{y_2} = - 9$ we get,
Centre $ = \left\{ {\dfrac{{ - 4 + 8}}{2},\dfrac{{7 - 9}}{2}} \right\}$
$ \Rightarrow = \left( {\dfrac{4}{2},\dfrac{{ - 2}}{2}} \right)$
Centre$ = \left( {2, - 1} \right)$
To calculate the radius use the distance formula, $r = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $ where $\left( {{x_1},{y_1}} \right)$and$\left( {{x_2},{y_2}} \right)$are two points.
The two points here are the centre $\left( {2, - 1} \right)$and the end point$\left( { - 4,7} \right)$,
$r = \sqrt {{{\left( { - 4 - 2} \right)}^2} + {{\left( {7 - ( - 1)} \right)}^2}} $
On simplifying
$ \Rightarrow = \sqrt {{{\left( { - 6} \right)}^2} + {{\left( 8 \right)}^2}} $
Square of the numbers under the root
$ = \sqrt {36 + 64} $
Adding the numbers under the root
$ \Rightarrow r = \sqrt {100} $
$\therefore $$r = 10$
The standard form of the equation of a circle ${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$ where $\left( {a,b} \right)$ are the coordinates of the centre and $r$ the radius. The centre and radius is $\left( {2, - 1} \right)$and $r = 10$
The equation of the circle can now be written,
${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$
Putting the values of the centre and radius
$ \Rightarrow {\left( {x - 2} \right)^2} + {\left( {y - ( - 1)} \right)^2} = {10^2}$
On simplified,
$ \Rightarrow {\left( {x - 2} \right)^2} + {\left( {y + 1} \right)^2} = 100$

Hence the equation of the circle is ${\left( {x - 2} \right)^2} + {\left( {y + 1} \right)^2} = 100$
It also can be written as, ${x^2} + {y^2} - 4x + 2y - 95 = 0$

Note: We can also use the formula to find the equation of a circle with $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$extremities of one of the diameters of the circle is $\left( {x - {x_1}} \right)\left( {x - {x_2}} \right) + \left( {y - {y_1}} \right)\left( {y - {y_2}} \right) = 0$
Here the endpoints of its diameter at $\left( { - 4,7} \right)$ and $\left( {8, - 9} \right)$
${x_1} = - 4\,,\,{x_2} = 8\,,\,{y_1} = 7\,,\,{y_2} = - 9$
Substitute this value in the formula $\left( {x - {x_1}} \right)\left( {x - {x_2}} \right) + \left( {y - {y_1}} \right)\left( {y - {y_2}} \right) = 0$
$ \Rightarrow \left( {x - ( - 4)} \right)\left( {x - 8} \right) + \left( {y - 7} \right)\left( {y - ( - 9)} \right) = 0$
$ \Rightarrow \left( {x + 4} \right)\left( {x - 8} \right) + \left( {y - 7} \right)\left( {y + 9} \right) = 0$
$ \Rightarrow {x^2} - 8x + 4x - 32 + {y^2} + 9y - 7y - 63 = 0$
$ \Rightarrow {x^2} + {y^2} - 4x + 2y - 95 = 0$