### Related Formula (x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$ This is the equation of a circle passing through diametrically opposite endpoints (x_1, y_1)$(x_1, y_1)$ and (x_2, y_2)$(x_2, y_2)$. ### Core Logic Look at the three known points: A(1, 0)$A(1, 0)$, B(0, 1)$B(0, 1)$, and O(0, 0)$O(0, 0)$. The vectors vecOA$\vec{OA}$ and vecOB$\vec{OB}$ align with the x and y axes, meaning angle AOB = 90^circ$\angle AOB = 90^{\circ}$. Since the angle subtended by AB$AB$ at a point O$O$ on the circle is 90^circ$90^{\circ}$, the segment AB$AB$ must be the diameter of the circle. ### Step 1: Finding the Circle Equation Using the diametric form for points A(1,0)$A(1,0)$ and B(0,1)$B(0,1)$: (x-1)(x-0) + (y-0)(y-1) = 0$(x-1)(x-0) + (y-0)(y-1) = 0$ x^2 - x + y^2 - y = 0$x^2 - x + y^2 - y = 0$ ### Step 2: Solving for k Since the point (2k, 3k)$(2k, 3k)$ also lies on this circle, substitute x = 2k$x = 2k$ and y = 3k$y = 3k$ into the equation: (2k)^2 - (2k) + (3k)^2 - (3k) = 0$(2k)^2 - (2k) + (3k)^2 - (3k) = 0$ 4k^2 - 2k + 9k^2 - 3k = 0$4k^2 - 2k + 9k^2 - 3k = 0$ 13k^2 - 5k = 0$13k^2 - 5k = 0$ k(13k - 5) = 0$k(13k - 5) = 0$ ### Step 3: Finding valid points This gives two possible values for k$k$: k = 0$k = 0$ or k = frac513$k = \frac{5}{13}$. If k=0$k=0$, the fourth point becomes (0,0)$(0,0)$, which is already listed. The question specifies four *distinct* points. Thus, k = frac513$k = \frac{5}{13}$. ### Pattern Recognition Whenever (0,0)$(0,0)$, (a,0)$(a,0)$, and (0,b)$(0,b)$ lie on a circle, they form a right triangle at the origin. The hypotenuse connecting (a,0)$(a,0)$ and (0,b)$(0,b)$ is ALWAYS the diameter. You can immediately write the circle's equation as x^2 + y^2 - ax - by = 0$x^2 + y^2 - ax - by = 0$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Circles