Question

please solve this its quite urgent​

Answer 1

We can easily observe that point O lies on AB which is perpendicular to OP. It is also marked Y and X so that we can understand that they are coordinate axes.Point O lies in the intersection points of both the coordinate axes.Obviously, this point known as origin should have coordinates (0,0).

The first question can be cracked by simple observation itself.

Length of OP = $\sqrt{(0-0)^2+(0-4)^2}$

= 4 units.

Now,let C be the centre of the circle.Join CP.

Let the radius of circle be R.

CP = R,

AB = 2R [ Diameter ]

Let the coordinate of B be (x,0) since the point lies on the x-axis itself.

Length of OA = $\sqrt{(-8-0)^2+(0-0)^2}$

= 8 units.

Length of OB = $\sqrt{(x-0)^2+(0-0)^2}$

= x units.

From figure , |OA| + |OB| = |AB|

⇒ 8 + x = 2R

$\implies R=\dfrac{x}{2}+4 \: ------>(1)$

Also from figure,

⇒ |OC| + |OB| = |CB|

⇒ |OC| = |CB| - |OB|

⇒ |OC| = R - x [ CB is the dam.n radius ]

Now, in Δ OPC,

OC² + OP² = PC² [ Pythagoras theorem ]

$\implies (R-x)^2+4^2=R^2\\\\\implies (\dfrac{x}{2}+4-x)^2+16=(\dfrac{x}{2}+4)^2[\textsf{See equation (1)}]\\\\\implies (4-\dfrac{x}{2})^2+16=(\dfrac{x}{2}+4)^2\\\\\implies 16+\dfrac{x^2}{4}-4x+16=16+\dfrac{x^2}{4}+4x\\\\\implies 8x=16\\\\\implies \boxed{x=2}$

$\boxed{\mathcal{ANSWERS}}$

(a) OP = 4 units.

(b) B has coordinate of (2,0).