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# ${\text{Find the equation of a circle with centre (2,2) and passes through the point (4,5)}}{\text{.}} \\ \\$

Last updated date: 17th Jul 2024
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${\text{Given, Centre C(2,2) and P(4,5) is a point on the circle}} \\ {\text{As, we know that the equation of circle which is passing through the centre}} \\ {\text{with coordinates (}}h,k{\text{) and radius '}}r{\text{' is }}{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2} \\ {\text{Therefore, putting Centre C(}}h = 2,k = 2{\text{) in above equation}} \\ \Rightarrow {\left( {x - 2} \right)^2} + {\left( {y - 2} \right)^2} = {r^2}{\text{ }}.............{\text{(1)}} \\ {\text{Since, P(4,5) is the passing point so this point will satisfy equation (1)}} \\ {\text{i}}{\text{.e}}{\text{. Put }}x = 4{\text{ and }}y = 5 \\ \Rightarrow {\left( {4 - 2} \right)^2} + {\left( {5 - 2} \right)^2} = {r^2} \Rightarrow {2^2} + {3^2} = {r^2} \Rightarrow {r^2} = 13 \\ {\text{Now put the value of }}{r^2}{\text{ in equation (1), we have}} \\ \Rightarrow {\left( {x - 2} \right)^2} + {\left( {y - 2} \right)^2} = 13 \\ {\text{This is the equation of the required circle}}{\text{.}}$