Answer

Verified

400.8k+ views

**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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Difference Between Plant Cell and Animal Cell

Which are the Top 10 Largest Countries of the World?

10 examples of evaporation in daily life with explanations

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE