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Find the equation of a circle with centre (2,2) and pass through the point (4,5).

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Last updated date: 21st Feb 2024
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IVSAT 2024
Answer
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Hint: Consider the equation of circle $(x - h)^2 +(y - k)^2 = {r^2}$ and substitute centre C (h = 2,k = 2) and the point (x = 4,y=5) to find the equation of circle. 


Complete step by step answer:

Given, Centre C(2,2) and P(4,5) is a point on the circle.

As, we know that the equation of circle which is passing through the centre with coordinates (h,k) and radius 'r' is 

$(x - h)^2 +(y - k)^2 = {r^2} $

Therefore, putting Centre C (h = 2,k = 2) in above equation,

$\Rightarrow {\left( {x - 2} \right)^2} + {\left( {y - 2} \right)^2} = {r^2}{\text{ }}.............{\text{(1)}} $

Since, P(4,5) is the passing point so this point will satisfy equation (1) i.e. Put x = 4 and y = 5 

$\Rightarrow {\left( {4 - 2} \right)^2} + {\left( {5 - 2} \right)^2} = {r^2} \Rightarrow {2^2} + {3^2} = {r^2} \Rightarrow {r^2} = 13 $

Now put the value of $r^2$ in equation (1), we have

$\Rightarrow {\left( {x - 2} \right)^2} + {\left( {y - 2} \right)^2} = 13 $

This is the equation of the required circle.

Note - In these types of problems the given data should be used in one of the general forms of equation of circle and the passing point helps to find the missing parameter (here it is radius).


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