Kerala KEAM 5 June 2024 Solved Paper

We are given the equation of the circle in the form: $$(x - 3)(x + 1) + (y - 1)(y + 3) = 0$$Let's expand the terms in the equation:1. For the first part: $$(x - 3)(x + 1) = x^2 + x - 3x - 3 = x^2 - 2x - 3$$2. For the second part: $$(y - 1)(y + 3) = y^2 + 3y - y - 3 = y^2 + 2y - 3$$Substituting these into the original equation: $$x^2 - 2x - 3 + y^2 + 2y - 3 = 0$$Simplify the equation: $$x^2 + y^2 - 2x + 2y - 6 = 0$$Now, we rearrange the equation into standard form for a circle: $$x^2 - 2x + y^2 + 2y = 6$$Next, we complete the square for both $x$ and $y$.- For $x^2 - 2x$, take half of -2, which is -1, square it to get 1. So, add and subtract 1.- For $y^2 + 2y$, take half of 2, which is 1, square it to get 1. So, add and subtract 1.Thus, we have: $$(x^2 - 2x + 1) + (y^2 + 2y + 1) = 6 + 1 + 1$$Simplifying: $$(x - 1)^2 + (y + 1)^2 = 8$$This is the standard equation of a circle with center $(h, k)$ and radius $\sqrt{8}$.So, the center of the circle is $(1, -1)$.Thus, the correct answer is option (E), $(1, -1)$. Quick Tip: To find the center of a circle from the equation $(x-h)(x+k)+(y-p)(y+q)=0$, expand the terms and simplify the equation into the standard form $(x-h)^2 + (y-p)^2 = r^2$.