Equation of circle inscribed in |x-a|+|y-b|=1 is

Question

Equation of circle inscribed in |x-a|+|y-b|=1 is (A) (x+a)2+(y+b)2=2 (B) (x-a)2+(y-b)2=(1/2) (C) (x-a)2+(y-b)2=(1/√2) (D) (x-a)2+(y-b)2=1

Tardigrade Admin · Accepted Answer

|x-a|+|y-b|=1 is square with center at (a, b) which is center of circle also. Distance between two parallel sides of square is √2. So, radius of the required circle is (1/√2). Hence equation is (x-a)2+(y-b)2=(1/2)

Q. Equation of circle inscribed in $∣ x - a ∣ + ∣ y - b ∣ = 1$ is

$(x + a)^{2} + (y + b)^{2} = 2$

$(x - a)^{2} + (y - b)^{2} = \frac{1}{2}$

$(x - a)^{2} + (y - b)^{2} = \frac{1}{2}$

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

$∣ x - a ∣ + ∣ y - b ∣ = 1$ is square with center at $(a, b)$ which is center of circle also.
Distance between two parallel sides of square is $2$ .
So, radius of the required circle is $\frac{1}{2}$ .
Hence equation is $(x - a)^{2} + (y - b)^{2} = \frac{1}{2}$

Q. Equation of circle inscribed in ∣x−a∣+∣y−b∣=1 is

(x+a)2+(y+b)2=2

(x−a)2+(y−b)2=21​

(x−a)2+(y−b)2=2​1​

(x−a)2+(y−b)2=1

∣x−a∣+∣y−b∣=1 is square with center at (a,b) which is center of circle also. Distance between two parallel sides of square is 2​. So, radius of the required circle is 2​1​. Hence equation is (x−a)2+(y−b)2=21​

Q. Equation of circle inscribed in $∣ x - a ∣ + ∣ y - b ∣ = 1$ is

$(x + a)^{2} + (y + b)^{2} = 2$

$(x - a)^{2} + (y - b)^{2} = \frac{1}{2}$

$(x - a)^{2} + (y - b)^{2} = \frac{1}{2}$

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

$∣ x - a ∣ + ∣ y - b ∣ = 1$ is square with center at $(a, b)$ which is center of circle also.
Distance between two parallel sides of square is $2$ .
So, radius of the required circle is $\frac{1}{2}$ .
Hence equation is $(x - a)^{2} + (y - b)^{2} = \frac{1}{2}$