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Question: A circle of Radius \( 2 \) unit lies in the First Quadrant and touches both the axes of coordinates,...

A circle of Radius 22 unit lies in the First Quadrant and touches both the axes of coordinates, equation of circle with center at (6,5)(6,5) and touching the above circle externally is
A−A - (x−6)2+(y−5)2=9{(x - 6)^2} + {(y - 5)^2} = 9
B−B - x2+y2−10x−12y+52=0{x^2} + {y^2} - 10x - 12y + 52 = 0
C−C - x2+y2−12x−10y+32=0{x^2} + {y^2} - 12x - 10y + 32 = 0
D−D - x2+y2+12x−10y−32=0{x^2} + {y^2} + 12x - 10y - 32 = 0

Explanation

Solution

Hint : In order to solve numerical geometry we need to know each and every property along with all the formulae. For this sum we will have to use the property Sum of the radius=Distance between their centers. We have used this because it is given circle touch externally. Since this is a circle question we will use (x−h)2+(y−k)2=c2{(x - h)^2} + {(y - k)^2} = {c^2} as the general equation

** Complete step-by-step answer** :
Let the centre of the given circle be (h,k)(h,k)
Given that circle lies in the First Quadrant and touches both the axes of coordinates.
∴h=k=2\therefore h = k = 2
The equation of the circle formed with Radius 22 units and value of h,k=2h,k = 2 is
(x−2)2+(y−2)2=22...............(1){(x - 2)^2} + {(y - 2)^2} = {2^2}...............(1)
Now, Let the radius of the required circle be RR .
∴\therefore Equation of Circle with coordinates (6,5)(6,5) and radius as RR is
(x−6)2+(y−5)2=R2...............(2){(x - 6)^2} + {(y - 5)^2} = {R^2}...............(2)
Given that Circle 11 and Circle 22 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
(x1−a)2+(y1−b)2\sqrt {{{({x_1} - a)}^2} + {{({y_1} - b)}^2}} , where (x1,y1)&(a,b)({x_1},{y_1})\& (a,b) are coordinates of 22 points between which we have to find the distance.
∴\therefore ⇒R+2=(6−2)2+(5−2)2\Rightarrow R + 2 = \sqrt {{{(6 - 2)}^2} + {{(5 - 2)}^2}}
⇒R+2=42+32\Rightarrow R + 2 = \sqrt {{4^2} + {3^2}}
⇒R+2=25\Rightarrow R + 2 = \sqrt {25}
Since Radius Is always Positive we will take only positive values of the root.
⇒R+2=5\Rightarrow R + 2 = 5
⇒R=3\Rightarrow R = 3
Substituting value of RR in Equation2Equation2 ,we get the Equation of the Required Circle is
⇒(x−6)2+(y−5)2=32\Rightarrow {(x - 6)^2} + {(y - 5)^2} = {3^2}
⇒(x−6)2+(y−5)2=9\Rightarrow {(x - 6)^2} + {(y - 5)^2} = 9
Thus answer to the given solution is Option AA i.e. (x−6)2+(y−5)2=9{(x - 6)^2} + {(y - 5)^2} = 9
So, the correct answer is “Option A”.

Note : It is advisable to learn distance formulae and general equations for figures like Parabola, Hyperbola, Ellipse, Circle. It would help to solve problems where there are more than 11 geometric figures involved for finding the equation for distance between the two figures.