Question

Assertion: The common chord of the circles ${{x}^{2}}+{{y}^{2}}-10x+16=0$ and ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$ is of maximum length if ${{r}^{2}}=34$.
Reason: The common chord of two circles is of maximum length if it passes through the centre of the circle with smaller radius.
A. Both assertion and reason are correct and reason is the correct explanation for assertion
B. Both assertion and reason are correct but reason is not correct explanation for assertion
C. Assertion is correct but reason is incorrect
D. Both assertion and reason are incorrect

Answer 1

Hint: If we want a common chord of maximum length then the common chord must be the diameter of the smaller circle (with smaller radius). Find the equation of the common chord between the two circles.

Formula used:
Equation of common chord :- ${{S}_{1}}-{{S}_{2}}=0$

Complete step by step answer:
There are two circles given to us whose equations are ${{x}^{2}}+{{y}^{2}}-10x+16=0$ and ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$, respectively.
It is claimed in the assertion that the length of the common chord of the two circles is maximum if ${{r}^{2}}=34$.
Now, suppose there are two circles of unequal radii. Then the two circles will have a common chord if the two meet or intersect at two different points. But if we want a common chord of maximum length then the common chord must be the diameter of the smaller circle (with smaller radius).
We know that the diameter of a circle passes through its centre. This means that the common chord must pass through the centre of the smaller circle.
Now, let us find the equation for the common chord of the two given circles.
For this write the equation of the circles in the form $S=0$.
Let the equation of the first circle be ${{S}_{1}}=0$ and the equation of the other circle be ${{S}_{2}}=0$.
Therefore,
 ${{S}_{1}}={{x}^{2}}+{{y}^{2}}-10x+16=0$ .…. (i).
${{S}_{2}}={{x}^{2}}+{{y}^{2}}-{{r}^{2}}=0$
Let the circle ${{S}_{1}}=0$ be the smaller circle. This means that the common chord is passing through the centre of this circle.
 We can write (i) as $\left( {{x}^{2}}-10x+25 \right)-25+{{y}^{2}}+16=0$
$\Rightarrow {{(x-5)}^{2}}+{{y}^{2}}=9$.
$\Rightarrow {{(x-5)}^{2}}+{{y}^{2}}={{3}^{2}}$
This means that the centre of the circle is (5,0) and its radius is 3.
The equation of the common chord is given as ${{S}_{1}}-{{S}_{2}}=0$.
Then means that equation of the common chord is $\left( {{x}^{2}}+{{y}^{2}}-10x+16 \right)-\left( {{x}^{2}}+{{y}^{2}}-{{r}^{2}} \right)=0$
$\Rightarrow -10x+16+{{r}^{2}}=0$ …. (ii)
Since the chord passes through the centre of the circle ${{S}_{1}}=0$, (5,0) satisfies the equation (ii).
Therefore, substitute $x=50$ in (ii).
$\Rightarrow -10(5)+16+{{r}^{2}}=0$
$\Rightarrow {{r}^{2}}=34$.
Therefore, we can say that the common chord of the circles ${{x}^{2}}+{{y}^{2}}-10x+16=0$ and ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$ is of maximum length if ${{r}^{2}}=34$.
And the reason is that the common chord of two circles is of maximum length if it passes through the centre of the circle with smaller radius.
This means that the assertion and the reason are both correct and the reason is the correct explanation for the assertion.

So, the correct answer is Option B.

Note: If you think why we considered the circle with equation ${{S}_{1}}=0$ as the smaller circle and the not the circle whose equation is ${{S}_{2}}=0$, then the reason for this that if we consider the other circle ${{S}_{2}}=0$ as the smaller circle then its diameter will also be small. As a result we will not achieve the maximum length of the common chord.