Q56 / 90 · JEE Main 2022 (24 Jun Shift 2) · Mathematics

Let a circle $C : {(x - h)}^{2} + {(y - k)}^{2} = r^{2}, k > 0$ , touch the $x$ -axis at $(1, 0)$ . If the line $x + y = 0$ intersects the circle $C$ at $P$ and $Q$ such that the length of the chord $P Q$ is $2$ , then the value of $h + k + r$ is equal to _____.

Correct answer: 7

Solution

Since the circle is touching $x$ -axis so, $k = r$ and $h = 1$

Now by diagram, $O P = r, P R = 1$

$O R = |\frac{r + 1}{\sqrt{2}}|$

$r^{2} = 1 + \frac{{(r + 1)}^{2}}{2}$

$2 r^{2} = 2 + r^{2} + 1 + 2 r$

$r^{2} - 2 r - 3 = 0$

$(r - 3) (r + 1) = 0$

$r = 3, - 1$

$h + k + r = 1 + 3 + 3$

$= 7$

Hence, the correct answer is 7.