Question

How do you find the equation of a circle passing through the point (4,3), (-2,-5) and (5,2)?

Answer 1

Hint: Here the question need to find the equation of circle by using three given coordinates, here we know that the equation of circle will satisfy all the given three points, here we have to use the line equation and then by sing it we can reach to the centre of the circle, and finally obtain the main equation by using centre and radius of the given circle.

Complete step by step answer:
The given question need to find the equation of circle which is passing through the given coordinates,
Let us first gives the name to the given coordinates:
On saying we get:
\[ \Rightarrow A = (4,3),\,B = ( - 2, - 5),C = (5,2)\]
Let us first find the midpoint of AB and BC:
Midpoint of AB \[ \ = \dfrac{{4 + ( - 2)}}{2},\dfrac{{3 + ( - 5)}}{2} = (1, - 1) = D\]
Midpoint of BC \[ \ = \dfrac{{ - 2 + (5)}}{2},\dfrac{{ - 5 + 2}}{2} = (1.5, - 1.5) = E\]
Now we have to find the equation of line passing through the points D and E and perpendicular to AB and BC respectively, on solving we get:
line passing to D and perpendicular to line AB
Line $ AB \to y - 3 = \dfrac{{ - 5 - 3}}{{ - 2 - 4}}x - 4 \to y - 3 = \dfrac{4}{3}x - 4 \to y = \dfrac{4}{3}x - 1 $
Perpendicular line equation:slope $= \dfrac{{ - 3}}{4}$ (negative slope of the line)
By using y = slope(x) + b(for perpendicular line equation)
Using the point D(1, - 1) we get
$\Rightarrow - 1 = \dfrac{{ - 3}}{4}(1) + b \to b = - 1 $
Hence equation of line $y = \dfrac{{ - 3}}{4}x - 1$
Now for line perpendicular to BC and passing through E we get:
Line passing to E and perpendicular to line BC
Line AB $\to y - 2 = \dfrac{{2 - ( - 5)}}{{5 - ( - 2)}}x - 5 \to y - 2 = \dfrac{7}{7}x - 5 \to y = 1x - 5 $
Perpendicular line equation:slope = - 1(negative slope of the line)
by using y = slope(x) + b(for perpendicular line equation)
Using the pont E(1.5, - 1.5)
$\Rightarrow - 1 = - 1(1) + b \to b = 0 $
Hence equation of line y = - 1x
Centre of the circle is the point of intersection of the above two lines, on solving we get:
Point of intersection of lines $y = \dfrac{{ - 3}}{4}x - 1\,and\,y = - x $
Putting y = - x in first line we get
\[ \Rightarrow - x = \dfrac{{ - 3}}{4}x - 1 \\
   \Rightarrow - 4x = - 3x - 1 \\
   \Rightarrow - x = - 1 \\
   \Rightarrow x = 1,\,so\,y = - 1 \\
   \]
Centre of circle is (1, - 1)

Radius of circle, on solving we get:
Radius of circle = distance between centre and any one coordinate of A,B and C
\[
      \Rightarrow \sqrt {{{(1 - 4)}^2} + {{( - 1 - 3)}^2}} = \sqrt {{{( - 3)}^2} + {{( - 4)}^2}} = \sqrt {25} = 5 \\
 \]
Final equation of the circle is:
\[
   \Rightarrow {(x - 1)^2} + {(y - ( - 1))^2} = 5^2 \\
   \Rightarrow {(x - 1)^2} + {(y + 1)^2} = 5^2 \\
 \]
This is the required equation for the circle.

Note: The given question can also be solved by plotting the coordinates on graph and then completing the point locus to a circle, and obtaining the centre accordingly, on the centre is achieved then by using any point we can get the radius and then equation for circle.