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Question

Answers

A) \[2\]

B) \[3\]

C) \[4\]

D) \[5\]

Answer
Verified

Hint: The general form of straight line \[y = mx + c\], where \[m\]be the slope of the straight line.

The general form of the circle passes through the centre \[(a,b)\] is, \[{(x - a)^2} + {(y - b)^2} = {r^2}\], where, \[r\] be the radius.

__Complete step by step answer:__

It is given that, \[ax + by = 10\] is the chord of minimum length of the circle \[{(x - 10)^2} + {(y - 20)^2} = 729\] and the chord passes through\[(5,15)\].

First, we will find the slope of the given chord \[ax + by = 10\]

To find the chord let us rewrite the equation of the chord as,

\[y = - \dfrac{a}{b}x + \dfrac{{10}}{b}\]

Let us Compare the above equation with the general form of straight line \[y = mx + c\] we get,

The slope of the chord \[ax + by = 10\] is \[ - \dfrac{a}{b}\].

That is $m = - \dfrac{a}{b}$

The general form of the circle passes through the centre \[(a,b)\] is\[{(x - a)^2} + {(y - b)^2} = {r^2}\],

Where, \[r\] be the radius.

Let us compare\[{(x - 10)^2} + {(y - 20)^2} = 729\] with the general form of circle we get,

The centre of the given circle is\[(10,20)\].

We know that, the slope of any straight line passes through the centre and meets the chord is \[1.\]

Again, any chord and the straight line passing through its centre is perpendicular to each other.

So, \[\dfrac{{ - a}}{b} \times 1 = - 1\], then we get,\[ - a = - b\]which can be written as \[a = b\]

Then \[ax + by = 10\]can be rewritten as \[ax + ay = 10\]

Now, the chord passes through the point \[(5,15)\]substitute the points in the above equation,

\[a \times 5 + a \times 15 = 10\]

Solving we get,

\[

20a = 10 \\

or,a = \dfrac{1}{2} \\

\]

Since,\[a = b\] ,\[b = \dfrac{1}{2}\]

So, substitute the value of \[a,b\] in \[4a + 2b\]we get,

\[4 \times \dfrac{1}{2} + 2 \times \dfrac{1}{2} = 2 + 1 = 3\]

__Hence, the correct option is (B)__ \[3\].

Note:

Consider, a straight line \[ax + by + c = 0\] passes through\[(p,q)\], then \[ap + bq + c = 0\] can also be used to solve the problem.

The general form of the circle passes through the centre \[(a,b)\] is, \[{(x - a)^2} + {(y - b)^2} = {r^2}\], where, \[r\] be the radius.

It is given that, \[ax + by = 10\] is the chord of minimum length of the circle \[{(x - 10)^2} + {(y - 20)^2} = 729\] and the chord passes through\[(5,15)\].

First, we will find the slope of the given chord \[ax + by = 10\]

To find the chord let us rewrite the equation of the chord as,

\[y = - \dfrac{a}{b}x + \dfrac{{10}}{b}\]

Let us Compare the above equation with the general form of straight line \[y = mx + c\] we get,

The slope of the chord \[ax + by = 10\] is \[ - \dfrac{a}{b}\].

That is $m = - \dfrac{a}{b}$

The general form of the circle passes through the centre \[(a,b)\] is\[{(x - a)^2} + {(y - b)^2} = {r^2}\],

Where, \[r\] be the radius.

Let us compare\[{(x - 10)^2} + {(y - 20)^2} = 729\] with the general form of circle we get,

The centre of the given circle is\[(10,20)\].

We know that, the slope of any straight line passes through the centre and meets the chord is \[1.\]

Again, any chord and the straight line passing through its centre is perpendicular to each other.

So, \[\dfrac{{ - a}}{b} \times 1 = - 1\], then we get,\[ - a = - b\]which can be written as \[a = b\]

Then \[ax + by = 10\]can be rewritten as \[ax + ay = 10\]

Now, the chord passes through the point \[(5,15)\]substitute the points in the above equation,

\[a \times 5 + a \times 15 = 10\]

Solving we get,

\[

20a = 10 \\

or,a = \dfrac{1}{2} \\

\]

Since,\[a = b\] ,\[b = \dfrac{1}{2}\]

So, substitute the value of \[a,b\] in \[4a + 2b\]we get,

\[4 \times \dfrac{1}{2} + 2 \times \dfrac{1}{2} = 2 + 1 = 3\]

Note:

Consider, a straight line \[ax + by + c = 0\] passes through\[(p,q)\], then \[ap + bq + c = 0\] can also be used to solve the problem.

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