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$ A - $ $ {(x - 6)^2} + {(y - 5)^2} = 9 $

$ B - $ $ {x^2} + {y^2} - 10x - 12y + 52 = 0 $

$ C - $ $ {x^2} + {y^2} - 12x - 10y + 32 = 0 $

$ D - $ $ {x^2} + {y^2} + 12x - 10y - 32 = 0 $

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Let the centre of the given circle be $ (h,k) $

Given that circle lies in the First Quadrant and touches both the axes of coordinates.

$ \therefore h = k = 2 $

The equation of the circle formed with Radius $ 2 $ units and value of $ h,k = 2 $ is

$ {(x - 2)^2} + {(y - 2)^2} = {2^2}...............(1) $

Now, Let the radius of the required circle be $ R $ .

$ \therefore $ Equation of Circle with coordinates $ (6,5) $ and radius as $ R $ is

$ {(x - 6)^2} + {(y - 5)^2} = {R^2}...............(2) $

Given that Circle $ 1 $ and Circle $ 2 $ touch each other externally, We have to use the property

Sum of the radius=Distance between their Centre

The distance between the centre can be calculated by using the distance formula which is given by

$ \sqrt {{{({x_1} - a)}^2} + {{({y_1} - b)}^2}} $ , where $ ({x_1},{y_1})\& (a,b) $ are coordinates of $ 2 $ points between which we have to find the distance.

$ \therefore $ $ \Rightarrow R + 2 = \sqrt {{{(6 - 2)}^2} + {{(5 - 2)}^2}} $

$ \Rightarrow R + 2 = \sqrt {{4^2} + {3^2}} $

\[ \Rightarrow R + 2 = \sqrt {25} \]

Since Radius Is always Positive we will take only positive values of the root.

\[ \Rightarrow R + 2 = 5\]

\[ \Rightarrow R = 3\]

Substituting value of $ R $ in $ Equation2 $ ,we get the Equation of the Required Circle is

$ \Rightarrow {(x - 6)^2} + {(y - 5)^2} = {3^2} $

$ \Rightarrow {(x - 6)^2} + {(y - 5)^2} = 9 $

Thus answer to the given solution is Option $ A $ i.e. $ {(x - 6)^2} + {(y - 5)^2} = 9 $