Radius of a Circle: Solving Negative Equations

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AI Thread Summary
The discussion revolves around understanding the negative signs in the equations related to the radius of a circle. The user is confused about the negative in the expression h-2k+2 and the use of a negative square root in the denominator. It is clarified that when finding perpendicular distances from the center of the circle to the tangents, the constant C in the tangent equation should be made positive. This adjustment helps in accurately determining the radius and center of the circle. Overall, the focus is on resolving the confusion regarding the signs in the equations for proper calculation.
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Homework Statement



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Homework Equations



Why is there negative on h-2k+2?and a denominator of negative square root of 5?
also in 2h-k-17, it has a negative sign but the denominator is not negative square root of 5? I really don't understand this part... please explain :frown:
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The Attempt at a Solution

 
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In this problem center of the circle lies between the two tangents. In such cases, to find the perpendicular distances, you have to make the constant C in the tangent positive.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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