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If the sum of the coefficients in the expansion of ( 1+2x^{}

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If the sum of the coefficients in the expansion of $(1 + 2 x)^{m}$ and $(2 + x)^{n}$ are respectively $6561$ and $243$ , then the position of the point $(m, n)$ with respect to circle $x^{2} + y^{2} - 4 x - 6 y - 32 = 0$

is on the circle

is outside the circle

is inside the circle

circle can not be fixed

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Levi Discussed

If the sum of the coefficients in the expansion of

(1 + 2 x)^{m}

and

(2 + x)^{n}

are respectively

6561

and

243

, then the position of the point

(m, n)

with respect to circle

x^{2} + y^{2} - 4 x - 6 y - 32 = 0

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

Substituting

x = 1

, we get

3^{m} = 6561

\Rightarrow 3^{m} = 3^{8}

\Rightarrow m = 8

And

3^{n} = 243

3^{n} = 3^{5}

\Rightarrow n = 5

Substituting the point

(m, n)

in the given equation of circle we get

8^{2} + 5^{2} - 4 (8) - 6 (5) - 32

= 64 + 25 - 32 - 30 - 32

= 25 - 30

= - 5

- 5 < 0

hence the point is inside the circle.

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Question Text	If the sum of the coefficients in the expansion of $(1 + 2 x)^{m}$ and $(2 + x)^{n}$ are respectively $6561$ and $243$ , then the position of the point $(m, n)$ with respect to circle $x^{2} + y^{2} - 4 x - 6 y - 32 = 0$
Answer Type	Text solution:1
Upvotes	150

If the sum of the coefficients in the expansion of (1+2x)m and (2+x)n are respectively 6561 and 243, then the position of the point (m,n) with respect to circle x2+y2−4x−6y−32=0

Text solutionVerified

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If the sum of the coefficients in the expansion of $(1 + 2 x)^{m}$ and $(2 + x)^{n}$ are respectively $6561$ and $243$ , then the position of the point $(m, n)$ with respect to circle $x^{2} + y^{2} - 4 x - 6 y - 32 = 0$