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Find the center of a circle given the three points on the circle as $\left( { - 3,5} \right)$, $(3,3)$, $(11,19)$ ?

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Last updated date: 28th Feb 2024
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IVSAT 2024
Answer
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Hint:The given question requires us to find the centre of a circle given the three point on the circle as: $\left( { - 3,5} \right)$, $(3,3)$, $(11,19)$. So, we have to find the centre of the circle with the information that is given to us in the question. We can do so by putting in the values of the coordinates of the three points lying on the circle in the standard equation of circle.

Complete step by step solution:
Let us consider the equation of a circle in standard form whose centre is at $\left( {h,k} \right)$ and radius is r units.
So, we have,${(x - h)^2} + {(y - k)^2} = {r^2} - - - - (1)$
Now point $( - 3,5)$lies on a circle, thus it satisfies the equation of circle in equation $\left( 1 \right)$.
Hence, ${( - 3 - h)^2} + {(5 - k)^2} = {r^2}$
Opening the whole squares and simplifying calculations, we get,
$ \Rightarrow 9 + {h^2} + 6h + 25 + {k^2} - 10k = {r^2}$
$ \Rightarrow {h^2} + {k^2} + 6h - 10k + 34 = {r^2} - - - - - (2)$
Also, point $(3,3)$ lies on circle, thus it satisfies the equation of circle in equation $\left( 1 \right)$
${(3 - h)^2} + {(3 - k)^2} = {r^2}$
Opening the whole squares and simplifying calculations, we get,
$ \Rightarrow 9 + {h^2} - 6h + 9 + {k^2} - 6k = {r^2}$
\[ \Rightarrow {h^2} + {k^2} - 6h - 6k + 18 = {r^2}\]
Equating with equation$(2)$, we get,
\[ \Rightarrow {h^2} + {k^2} - 6h - 6k + 18 = {h^2} + {k^2} + 6h - 10k + 34\]
\[ \Rightarrow 12h - 4k + 16 = 0\]
On dividing both sides with $4$, we get,
$3h - k + 4 = 0 - - - - (3)$
Also, point$(11,19)$ lies in circle, thus it satisfies the equation of circle in equation$(1)$
${(11 - h)^2} + {(19 - k)^2} = {r^2}$
Opening the whole squares and simplifying calculations, we get,
$ \Rightarrow $$121 + {h^2} - 22h + 361 + {k^2} - 38k = {r^2}$
$ \Rightarrow $${h^2} + {k^2} - 22h - 38k + 482 = {r^2}$
Equating with equation$(2)$,
${h^2} + {k^2} - 22h - 38k + 482 = {h^2} + {k^2} + 6h - 10k + 34$
$ \Rightarrow $$28h + 28k - 448 = 0$
On dividing both sides with $28$,
$ \Rightarrow $$h + k - 16 = 0 - - - - (4)$
Now solving equation $(3)$ and $(4)$, we get,
$h = 3$ and $k = 13$
Thus coordinates of center of circle are $(3,13)$

Note: The centre of a circle can be found easily if three points lying on the circle are given to us by substituting the coordinates of the points in the equation of the circle and solving the three equations obtained for the coordinates of the centre.