Equation of chord AB of circle $x^{2} + y^{2} = 2$ passing through $(2, 2)$ such that $P B / P A = 3$ , is given by

x = 3 y

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y - 2 = \sqrt{3} (x - 2)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = y

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $x = y$
Given, circle $x^{2} + y^{2} = 2$ and $A B$ is the chord
Point $P (2, 2)$ is coordinate such that $B A$ passes through $P$
$\frac{P B}{A} = 3 \Rightarrow P B = 3 P A$
$\Rightarrow B A = 2 P A$ and $A M = B M = P A$ ( $M$ is the midpoint of chord)
In $△ A O M$
$O M = \sqrt{(O A)^{2} - (A M)^{2}} = \sqrt{2 - (A M)^{2}} (1)$
In $△ P O M$
$O M = \sqrt{(O P)^{2} - (P M)^{2}} = \sqrt{((2 - 0)^{2} + (2 - 0)^{2}) - (2 A M)^{2}} O M = \sqrt{8 - 4 (A M)^{2}} (2)$
From $(1)$ and $(2)$
$\sqrt{2 - (A M)^{2}} = \sqrt{8 - 4 (A M)^{2}} 2 - (A M)^{2} = 8 - 4 (A M)^{2} 3 (A M)^{2} = 6 A M = \sqrt{2} = r a d i u s$
Hence, $A M$ passes through origin and $A B$ is diameter
Equation of chord is
$(y - 0) = (\frac{2 - 0}{2 - 0}) (x - 0) y = x$

Suggest Corrections

Similar questions

Equation of chord AB of circle $x^{2} + y^{2} = 2$ passing through P(2, 2) such that $\frac{P B}{P A} = 3$ , is given by

P (1, 1)

x^{2} + y^{2} + 2 x + 2 y - 8 = 0.

A B

P .

\frac{P A}{P B} = \frac{\sqrt{5} - 2}{\sqrt{5} + 2},

A B

A B

x^{2} + y^{2} = r^{2}

(1, 1)

\frac{P B}{P A} = \frac{\sqrt{2} + r}{\sqrt{2} - r} \cdot (0 < r < \sqrt{2})

x^{2} + y^{2} - 2 x + 4 y = 0

Join BYJU'S Learning Program