Let C be a circle with radius sqrt{10} units | vedclass

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(B) The equation of the circle $C$ is $x^2+y^2=10$. The radius $R = \sqrt{10}$.For chord $PQ$: The line is $x+y-2=0$. The perpendicular distance $d_1$

Vedclass · Accepted Answer

$3-\sqrt{2}$

Vedclass · Answer

(B) The equation of the circle $C$ is $x^2+y^2=10$. The radius $R = \sqrt{10}$.For chord $PQ$: The line is $x+y-2=0$. The perpendicular distance $d_1$ from the origin $(0,0)$ to the line $PQ$ is $d_1 = \frac{|0+0-2|}{\sqrt{1^2+1^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.For chord $MN$: The length of the chord is $2$ units. Let $A$ be the midpoint of $MN$. Then $AN = \frac{MN}{2} = 1$. In $\Delta OAN$,$OA^2 + AN^2 = ON^2$,where $ON$ is the radius $R = \sqrt{10}$.$OA^2 + 1^2 = (\sqrt{10})^2 \implies OA^2 = 9 \implies OA = 3$. Thus,the perpendicular distance $d_2$ from the origin to the chord $MN$ is $3$.Since both chords $PQ$ and $MN$ have the same slope $-1$,they are parallel. The distance between two parallel chords is $|d_1 \pm d_2|$.Distance $= |3 \pm \sqrt{2}|$.Thus,the possible distances are $3+\sqrt{2}$ or $3-\sqrt{2}$.Comparing with the given options,$3-\sqrt{2}$ is the correct choice.

Let $C$ be a circle with radius $\sqrt{10}$ units and centre at the origin. Let the line $x+y=2$ intersect the circle $C$ at the points $P$ and $Q$. Let $MN$ be a chord of $C$ of length $2$ units and slope $-1$. Then,the distance (in units) between the chord $PQ$ and the chord $MN$ is

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