Question

if the line $x + y = 6$ and $x + 2y = 4$ are the diameter of the circle, and circle passes through point $(2,6)$ then find its equation.

Answer 1

Hint Given, the line $x + y = 6$ and $x + 2y = 4$ are the diameter of the circle, and circle passes through point $(2,6)$. We have to find the equation of a given circle. First, we will solve two given equations of line to find the centre of the circle. Using centre and passing points we will find the radius of the circle. Then we will find the required equation of the circle by putting centre and radius in the standard equation of circle.

Formula Used:
 ${\left( {x - k} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$

Complete step by Step solution: The location of a circle in the Cartesian plane is represented by a circle equation. We can write the equation for a circle if we know the location of the circle's centre and how long its radius is. All of the points on the circle's circumference are represented by the circle equation.
The group of points whose separation from a fixed point has a constant value are represented by a circle. The radius of the circle, abbreviated r, is a constant that describes this fixed point, which is known as the circle's centre. The standard equation of circle is ${\left( {x - k} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$ having centre $\left( {h,k} \right)$ and radius $r$
Given, the line $x + y = 6$ and $x + 2y = 4$ are the diameter of the circle, and the circle passes through point $(2,6)$.
$x + y = 6$ …… (1)
$x + 2y = 4$ …… (2)
Subtracting equation (1) from equation (2)
We get,
$2y - y = 4 - 6$
$y = - 2$
Putting in the equation (1)
$x + ( - 2) = 6$
$x = 6 + 2$
$x = 8$
Hence the centre of circle is $(8, - 2)$
Circle is passing through $\left( {2,6} \right)$
radius $= \sqrt {{{\left( {8 - 2} \right)}^2} + {{\left( {6 - \left( { - 2} \right)} \right)}^2}} $
$ = \sqrt {{6^2} + {{( 8)}^2}} $
$ = \sqrt {36 + 64} $
$ = 10$
Standard equation of circle is ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$ having centre at $(h,k)$ and radius $r$.
${\left( {x - 8} \right)^2} + {\left( {y - \left( { - 2} \right)} \right)^2} = {10^2}$
${x^2} + 64 - 16x + {y^2} + 4 + 4y = 100$
${x^2} + {y^2} - 16x + 4y + 68 = 100$
${x^2} + {y^2} - 16x + 4y + 68 - 100 = 0$
${x^2} + {y^2} - 16x + 4y - 32 = 0$
Hence, the required equation of circle is ${x^2} + {y^2} - 16x + 4y - 32 = 0$

Hint: Should understand the problem to gain the information on how we can solve the question. The standard equation of circle is ${\left( {x - k} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$ having centre $\left( {h,k} \right)$ and radius $r$ should be used in solution to get the required equation of circle. Students should solve questions carefully to avoid any mistakes.