If the line $x cos α + y sin α = p$ represents the common chord $A P Q B$ of the circles $x^{2} + y^{2} = a^{2}$ and $x^{2} + y^{2} = b^{2} (a > b)$ as shown in the figure, then $A P$ is equal to

\sqrt{a^{2} + p^{2}} + \sqrt{b^{2} + p^{2}}

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\sqrt{a^{2} - p^{2}} + \sqrt{b^{2} - p^{2}}

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\sqrt{a^{2} - p^{2}} - \sqrt{b^{2} - p^{2}}

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\sqrt{a^{2} + p^{2}} - \sqrt{b^{2} + p^{2}}

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Solution

The correct option is C $\sqrt{a^{2} - p^{2}} - \sqrt{b^{2} - p^{2}}$
The given circles are concentric with centre at $(0, 0)$

The length of the perpendicular from $(0, 0)$ on the given line is $p$ .

Let $O L = p$ , then

$A L = \sqrt{(O A)^{2} - (O L)^{2}} = \sqrt{a^{2} - p^{2}}$

and $P L = \sqrt{(O P)^{2} - (O L)^{2}} = \sqrt{b^{2} - p^{2}}$

$\Rightarrow A P = \sqrt{a^{2} - p^{2}} - \sqrt{b^{2} - p^{2}}$

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Similar questions

If the straight line x cos $α$ + y sin $α$ = p touches the curve $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1,$ then prove that $a^{2} c o s^{2} α + b^{2} s i n^{2} α = p^{2}$ .

Q. If the straight line

x cos α + y sin α = p

touches the curve

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1,

then prove that

a^{2} {cos}^{2} α + b^{2} {sin}^{2} α = p^{2}

Q. If the straight line

x cos α + y sin α = p

touches the curve

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

, then prove that

a^{2} {cos}^{2} α - b^{2} {sin}^{2} α = p^{2}

Q. If the line

x cos α + y sin α = p

is a tangent to the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1,

then prove that

a^{2} {cos}^{2} α + b^{2} {sin}^{2} α = p^{2}

Q. If

P_{1}

and

P_{2}

are the perpendiculars from any point on the hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

on its asymptotes, then :

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If the line xcosα+ysinα=p represents the common chord APQB of the circles x2+y2=a2 and x2+y2=b2(a>b) as shown in the figure, then AP is equal to

The correct option is C √a2−p2−√b2−p2The given circles are concentric with centre at (0,0) The length of the perpendicular from (0,0) on the given line is p. Let OL=p, then AL=√(OA)2−(OL)2=√a2−p2and PL=√(OP)2−(OL)2=√b2−p2 ⇒AP=√a2−p2−√b2−p2

If the line $x cos α + y sin α = p$ represents the common chord $A P Q B$ of the circles $x^{2} + y^{2} = a^{2}$ and $x^{2} + y^{2} = b^{2} (a > b)$ as shown in the figure, then $A P$ is equal to