Let a circle C: (x-h)^{2} + (y-k)^{2} = r^{2} | vedclass

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(D) Since the circle touches the $x$-axis at $(1, 0)$,the center of the circle is $(h, k) = (1, r)$ and the radius is $r$. The distance from the cente

Vedclass · Accepted Answer

$7$

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(D) Since the circle touches the $x$-axis at $(1, 0)$,the center of the circle is $(h, k) = (1, r)$ and the radius is $r$. The distance from the center $(1, r)$ to the line $x + y = 0$ is given by $d = \frac{|1 + r|}{\sqrt{1^{2} + 1^{2}}} = \frac{|r + 1|}{\sqrt{2}}$. Since the length of the chord $PQ$ is $2$,the half-length is $1$. Using the Pythagorean theorem in the triangle formed by the radius,the distance $d$,and the half-chord,we have $r^{2} = d^{2} + 1^{2}$. Substituting the value of $d$,we get $r^{2} = \left(\frac{r + 1}{\sqrt{2}}\right)^{2} + 1$. $r^{2} = \frac{(r + 1)^{2}}{2} + 1$. $2r^{2} = r^{2} + 2r + 1 + 2$. $r^{2} - 2r - 3 = 0$. $(r - 3)(r + 1) = 0$. Since $r > 0$,we have $r = 3$. Thus,$h = 1$,$k = 3$,and $r = 3$. The value of $h + k + r = 1 + 3 + 3 = 7$.

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 $PQ$ is $2$,then the value of $h + k + r$ is equal to

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