Question

Prove that the points (7, –9) and (11, 3) lie on a circle with centre at the origin. Also find its equation.

Answer 1

Every points on a circle are equidistant from the center of the circle. We can prove the given in the question by finding the distances between each given points from the center (origin) and checking whether it is equal or not.

Distance between (7, -9) and (0, 0):

$\sqrt{(7 - 0)^2 + (-9-0)^2} \\ \\ \sqrt{7^2 + (-9)^2} \\ \\ \sqrt{49 + 81} \\ \\ \sqrt{130}$

Distance between (11, 3) and (0, 0):

$\sqrt{(11 - 0)^2 + (3-0)^2} \\ \\ \sqrt{11^2 + 3^2} \\ \\ \sqrt{121 + 9} \\ \\ \sqrt{130}$

Here, both are equal.

∴ Both lie on a circle with center at the origin.

Now let me write the equation.

Let (x, y) be a point on the circle.

Consider (x, y) and the center (0, 0).

We got that the radius is $\sqrt{130}$.

∴

$\sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{130} \\ \\ \sqrt{x^2 + y^2} = \sqrt{130} \\ \\ x^2 + y^2 = 130 \\ \\ x^2 + y^2 - 130 = 0$

So the equation is x² + y² - 130 = 0

Thank you. Have a nice day. :-)

#adithyasajeevan

Answer 2

Answer:

Step-by-step explanation: