Question

If \[ax + by = 10\] is the chord of minimum length of the circle \[{(x - 10)^2} + {(y - 20)^2} = 729\] and the chord passes through \[(5,15)\]then the value of \[4a + 2b\] is
A) \[2\]
B) \[3\]
C) \[4\]
D) \[5\]

Answer 1

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.