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Tanuj
Tanuj
15:20
Hey man , I have a doubt regarding this one question involving circles.
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Akiva Weinberger
Akiva Weinberger
Answer is B. $k\geqslant\frac12$
Tanuj
Tanuj
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@AkivaWeinberger I don't get why should $D$ should be greater than or equal to 0 for that equation.
i mean why does it need to have two real or same roots ?
Akiva Weinberger
Akiva Weinberger
$h^2+2h+(-2k+2)=0$ is a quadratic equation in $h$
Tanuj
Tanuj
@AkivaWeinberger yes.
But why do I need to prove it has two or atleast one root ?
Akiva Weinberger
Akiva Weinberger
$k$ is the $y$-coordinate of the center of the circle
Tanuj
Tanuj
15:25
@AkivaWeinberger for the circle to be a real circle ?
Akiva Weinberger
Akiva Weinberger
Yeah
If $k$ were chosen such that the quadratic had no real solutions, then $h$ would be imaginary and there would be no circle with center $(h,k)$
Tanuj
Tanuj
cool. But how would it strike my mind that to find a condition for $k$ I'll have to relate the discriminant
Akiva Weinberger
Akiva Weinberger
I would have solved it differently, to be honest
Tanuj
Tanuj
@AkivaWeinberger how did you solve it ? You didn't take a minute and came up with the correct answer.
Akiva Weinberger
Akiva Weinberger
Well if I needed to do it algebraically
From $(-1-h)^2+(1-k)^2=k^2$
@Tanuj I didn't solve it algebraically
Just think about the geometry, draw a picture if you need to
Tanuj
Tanuj
15:30
okay
Akiva Weinberger
Akiva Weinberger
But if I had to do it algebraically, the above is $k^2-(1-k)^2=(-1-h)^2$
$k^2-(1-k)^2\ge0$
Tanuj
Tanuj
@AkivaWeinberger how do i interpret that from the graph ?
Akiva Weinberger
Akiva Weinberger
Expand, $2k-1\ge0$, solve, $k\ge\frac12$
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Here's a crappy picture @Tanuj
Tanuj
Tanuj
What does it tell me ?
Akiva Weinberger
Akiva Weinberger
You can't get a circle to touch the point and be tangent to the x-axis if its center is less than ½ away from the x-axis
(where the distance from the x-axis to the point is 1)
The blue things are meant to be circles
Tanuj
Tanuj
15:37
@AkivaWeinberger ahhh got it.
@AkivaWeinberger Thanks so much for giving me your precious time :)

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