Question

Assertion :$C_1:{(x-3)}^2+{(y-4)}^2=a^2$

$C_1$, touches the axis of $x$ if $a=4$

Reason: $C_1$, touches the axis of $y=x$ if $a=3$

• 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 the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

Let the circle touch the x axis at $P=(x,0)$

If the centre of the circle lies at $C=(3,4)$

Then $P{C}^2=a^2$

$(x-3{)}^2=a^2-16$

$x=3\pm \sqrt{a^2-16}$

Now $x$ axis is a tangent to the circle at the point $(x,0)$

Hence, ${\frac{dy}{dx}}_{x,0}=0$

$2(x-3)+2(y-4)\frac{dy}{dx}=0$

Therefore, $2(x-3)=0$

$x=3$

$=3\pm \sqrt{a^2-16}$

Thus , $\sqrt{a^2-16}=0$

$\therefore a=4$

Hence assertion is correct.

Now if $y=x$ is a tangent to the circle, then

$(x-3{)}^2+(y-4{)}^2=a^2$
$(x-3{)}^2+(x-4{)}^2=a^2$
$2x^2-14x+25=a^2$

$2x^2-14x+25-a^2=0$ ...(i)

And ${\frac{dy}{dx}}_{x,y}=1$

$2(x-3)+2(y-4).{\frac{dy}{dx}}_{x,y}=0$
$2(x-3)+2(y-4)\left(1\right)=0$
$2x-6+2y-8=0$
$x+y=7$...(ii)

Since $(x,y)$ lies on $x=y$

Therefore

$2x=7$

$x=\frac{7}{2}$

Hence, $2x^2-14x+25-a^2=0$

$\frac{49}{2}-49+25-a^2=0$
$\frac{-49}{2}+25=a^2$
$\frac{1}{2}=a^2$
$\therefore a=\frac{1}{\sqrt{2}}$

Hence reason is incorrect.