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

Last updated date: 22nd Jul 2024
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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.

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).