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 $4a + 2b$ isA) $2$B) $3$C) $4$D) $5$

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.

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.