Question

From a point p outside the circle with centre o tangent segment pa and pb are drawn. if 1/oa^2+ 1/pa^2 = 1/16, compute ab

Answer 1

Answer: AB = 8

Step-by-step explanation:

The important part in the question is getting the figure right, please refer to the attached figure when referring to the answer.

Let the radius of the circle be 'r'.

∠PAO & ∠PBO is right angle because a tanget is always perpendicular to the radius.

Let ∠OPA be θ

In Δ PAO ,

Cot θ = $\frac{PA}{r}$

∴ PA = r Cot θ

Now, we have been given in the question as,

$\frac{1}{OA^2 }$ + $\frac{1}{PA^2}$ = $\frac{1}{16}$   , We know, OA = r and PA = r Cot θ

putting these values in above expression, we get

$\frac{1}{r^2}$ + $\frac{1}{r^2Cot^{2} theta }$ =  $\frac{1}{16}$

$\frac{1}{r^2}$  ( $\frac{Cot^{2} Theta + 1 }{Cot^{2} Theta }$ ) = $\frac{1}{16}$

We know, $Cot^{2} Theta + 1$ = $Cosec^{2} Theta$

$\frac{1}{r^2}$  ( $\frac{Cosec^2 Theta}{Cot^2 Theta}$ ) = $\frac{1}{16}$

Cot = $\frac{Cos}{Sin}$ and Cosec = $\frac{1}{sin}$ , putting these in above , we get,

$\frac{1}{r^2Cos^2 Theta}$ = $\frac{1}{16}$

∴ r cos θ = 4   ................... equation 1

In Δ PAM , ∠PMA = 90°

Tan θ = $\frac{x}{PM}$

∴ x = PM Tan θ    .................. equation 2

PM = PO - MO

In Δ PAO,

Sin θ = $\frac{r}{PO}$

∴ PO = r Cosec θ  ...................... equation 3

In Δ PAO,

∠OPA =   θ , ∠PAO = 90°

∴ ∠ POA = (90 - θ)

∴ Cos (90 - θ) = $\frac{MO}{r}$

we know, Cos (90 - θ) = SIn θ

∴ MO = r sin  θ   ........................ equation 4

∴ PM = PO - MO

Taking values of PO and MO from equations 3 and 4,

PM = r Cosec θ - r Sin θ

= r ( $\frac{1}{Sin Theta}$ - Sin θ )

= r ( $\frac{1 - Sin^2 Theta}{Sin Theta}$ )

= r ( $\frac{Cos^2 Theta}{Sin Theta}$ )

∴ PM = r ( $\frac{Cos^2 Theta}{Sin Theta}$ )   ....................... equation 5

As per equation 2 , we know

x = PM Tan θ

Substituting value of PM obtained in equation 5

x =  r ( $\frac{Cos^2 Theta}{Sin Theta}$ )  X Tan θ

= r ( $\frac{Cos^2 Theta}{Sin Theta}$ ) X $\frac{Sin Theta}{Cos Theta}$

x = r cos θ

But as per equation 1 , we know, r cos θ = 4

∴ x = 4

∴ AB = 2 X x = 2 X 4 = 8

∴ AB = 8