Question

the equation of the circle which passes through the point (4,6) and has its centre as 1,2​

Answer 1

SOLUTION :-

TO DETERMINE :-

The equation of the circle which passes through the point (4,6) and has its centre as (1,2)

EVALUATION :-

Let the equation of the circle with centre at (a, b) & radius r unit is

$\sf{ {(x - a)}^{2} + {(y - b)}^{2} = {r}^{2} \: } \: \: \: .....(1)$

Here it is given that the circle has centre at (1,2)

So a = 1, b = 2

Thus from Equation (1) we get

$\sf{ {(x - 1)}^{2} + {(y - 2)}^{2} = {r}^{2} \: } \: \: \: \: .....(2)$

Now this circle passes through the point (4,6)

So

$\sf{ {(4 - 1)}^{2} + {(6 - 2)}^{2} = {r}^{2} \: }$

$\implies \sf{ {3}^{2} + {4}^{2} = {r}^{2} \: }$

$\implies \sf{ {r}^{2} = {3}^{2} + {4}^{2} \: }$

$\implies \sf{ {r}^{2} = 9 + 16 \: }$

$\implies \sf{ {r}^{2} = 25 \: }$

From Equation (2) we get the equation of the circle as

$\sf{ {(x - 1)}^{2} + {(y - 2)}^{2} =25}$

FINAL ANSWER :-

The required equation of the circle is

$\boxed{ \sf{ \: \: {(x - 1)}^{2} + {(y - 2)}^{2} =25} \: \: }$

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