Question

3. If the points A (4, 3) and B (x. 5) are on the circle with centre 0 (2, 3), find the value of x​

Answer 1

Answer:

B(2, 5)

∴ x = 2

Step-by-step explanation:

We have two points on the circumference of the circle. When we join these to the centre of the circle, we get two radii that are equal to each other.

$\Longrightarrow \tt OB = OA$

$\Longrightarrow \tt \sqrt{(2-x)^2+(3-5)^2}=\sqrt{(2-4)^2+(3-3)^2}$

$\sf Squaring \ on \ both \ sides \ we \ get,$

$\Longrightarrow \tt \sqrt{(2-x)^2+(3-5)^2}^{ \ 2}=\sqrt{(2-4)^2+(3-3)^2}^{ \ 2}$

$\sf Roots \ and \ squares \ gets \ cancelled.$

$\Longrightarrow \tt (2-x)^2+(3-5)^2=(2-4)^2+(3-3)^2$

$\Longrightarrow \tt 4+x^2-4x+(-2)^2=(-2)^2+(0)^2$

$\Longrightarrow \tt 4+x^2-4x+4=4+0$

$\Longrightarrow \tt 4+x^2-4x+4-4=0$

$\Longrightarrow \tt 4+x^2-4x=0$

$\Longrightarrow \tt x^2-4x+4=0$

Sum ⇒ - 4

Product ⇒ 4

Splitting -4x = -2 × -2

$\Longrightarrow \tt x^2-2x-2x+4=0$

$\Longrightarrow \tt x(x-2)-2(x-2)=0$

$\Longrightarrow \tt (x-2)(x-2)=0$

Equating the zeroes with 0 we get,

$\Longrightarrow \tt x-2=0$

$\Longrightarrow \tt x=2$

∴ Therefore, the value of 'x' is 2.