Question

Equation of a circle touching the lines |x-1|+|y-3|=0 is

Answer 1

Answer:

⇒ x² + y² - 2x - 6y + 10 = 0

Step-by-step explanation:

Given :

Equation of a circle touching the lines | x - 1 | + | y - 3 | = 0

Solution :

We know that modulus of a number |x| (say x) is always positive,

Then,

Here in the given equation,

| x - 1 | + | y - 3 | = 0

There are two elements in modulus and nothing without modulus,

Sum of two or more positive numbers is always positive,

This means,

| x - 1 | + | y - 3 | = 0

⇒ | x - 1 | = - | y - 3 |

since one of the sides is inside modulus,

⇒ | x - 1 | ≥ 0

⇒ - | y - 3 | ≥ 0

Hence,

x - 1 = 0 & y - 3 = 0

∴ x = 1 & y = 3,.

∴ The required equation is :

⇒ | x - 1 | + | y - 3 | = 0

⇒ (x - 1) + (y - 3) = 0  (0 + 0 = 0)

Since x - 1 = 0 & y - 3 = 0

⇒ (x - 1)² + (y - 3)² = 0 (0² + 0² = 0)

⇒ ( x² - 2x + 1 ) + ( y² - 6y + 9) = 0

⇒ x² + y² - 2x - 6y + 10 = 0