Q. From a point P outside the circle with centre C.

Question

Q. From a point \( P \) outside the circle with centre at \( C \), tangents \( P A \) and \( P B \) are drawer such that they satisfy \( 4 P C=P A \cdot A C \). Then the length of chord \( A B \) is \( \alpha \) and the value of \( \alpha \) is \( x \) multiplied by itself single time find the value of \( \frac{x}{\sqrt{2}} \cdot(x \in R) \)

Answer 1

From point P, the lengths of the tangents PA and PB are equal. Let the length of the tangents be denoted as 't'. The distance from point P to the center C is denoted as 'd'. The radius of the circle is 'r'. By the Pythagorean theorem, we have:

1. PA² = PC² − AC² 

Thus, t² = d² − r²

2. The length of chord AB can be calculated using the formula: AB = 2⋅√PA² − AC²​

Substituting the value of PA, we get: AB = 2⋅√t² − r²​ 

Since t² = d² − r², we can substitute this into the equation for AB. Therefore, we have: AB = 2⋅ √(d² − r²) − r² ​ =2⋅√d² − 2r²​.

3. Given that the length of chord AB is represented as α, we can express it as α = 2⋅√d²−2r²​.

4. The problem states that the value of α is x multiplied by itself single time, which implies α = x². Therefore, we can equate: x² = 2⋅√d²−2r²​.

5. To find the value of 2/​√x​, we can simplify: x = √2⋅√d²−2r²​​ Thus, 2/√​x ​= √2⋅√d²−2r²/√2​​​ = √√d² − 2r² = (d²−2r²)^1/4.