Question

Which point does not lie on the circle centered at A(3, 1) and passing through the origin (0, 0)? A. J(2, -2) B. K(6, 0) C. L(4, -4) D. M(2, 4)

Answer 1

Answer:

The correct answer is C

Step-by-step explanation:

The given circle is centered at A(3, 1) and passes through the origin (0, 0).

The radius of this circle can be found using the distance formula;

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.

$\Rightarrow r=\sqrt{(3-0)^2+(1-0)^2}$.

$\Rightarrow r=\sqrt{(3)^2+(1)^2}$.

$\Rightarrow r=\sqrt{9+1}$.

$\Rightarrow r=\sqrt{10}$.

The equation of this circle is given by

$(x-a)^2+(y-b)^2=r^2$

Where $(a,b)$ is the centre.

We substitute the center and radius to obtain;

$(x-3)^2+(y-1)^2=(\sqrt{10})^2$

$(x-3)^2+(y-1)^2=10$

Option A

When we substitute (2,-2) into this equation we get;

$(2-3)^2+(-2-1)^2=10$

$1+9=10$

This is true

Option B

When we substitute K(6,0), we get,

$(6-3)^2+(0-1)^2=10$

$9+1=10$

This is also true

Option C

We substitute L(4,-4) to get;

$(4-3)^2+(-4-1)^2=10$

$1+25=10$

This is false. Hence L(4,-4) does not lie on the circle.

Option D

We substitute M(2,4)

$(2-3)^2+(4-1)^2=10$

$1+9=10$

This is also true.

The correct answer is C