Find the general equation of the circle with radius 5 and contains the points $A=(-8,0)$ and $B=(-4,-2)$.
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I don't know where to start – KelDScnd Apr 05 '17 at 18:04
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I don't know if this will help but my professor put a hint, "Hint: Equate radii. Or use the fact that D = h^2+k^2-r^2, then, systems." – KelDScnd Apr 05 '17 at 18:06
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There isn't a unique circle satisfying these properties. (To visualize this, plot any two points, and draw the two circles of equal radii that go through those two points.)
Mark Twain
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I don't know if this will help but my professor put a hint, "Hint: Equate radii. Or use the fact that D = h^2+k^2-r^2, then, systems." – KelDScnd Apr 05 '17 at 17:58
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Ross: Yes, but the question was to find the equation of the circle. – Mark Twain Apr 05 '17 at 18:09
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Hint: if the center is $(x,y)$, the statement that $(-8,0)$ is at a distance $5$ from it means $(x-(-8))^2+(y-0)^2=5^2$. You can get a second equation from the other point. Solve them simultaneously. You should find two points $(x,y)$ that satisfy the equations. Write the equation for a radius $5$ circle around each one.
Ross Millikan
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I don't get it, why would it be possible to put (-8,0) to the place of (h,k) if it isn't the points of the center of the circle? – KelDScnd Apr 05 '17 at 18:03
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You want the center to be $5$ units from $(-8,0)$, so we draw a circle of radius $5$ around $(-8,0)$ and say the center must be somewhere on the circle. – Ross Millikan Apr 05 '17 at 18:08
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Oh i get what your saying now, but still i'm confused on how to solve this question. – KelDScnd Apr 05 '17 at 18:11
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Can you write the corresponding equation based on $(-4,-2)$? That gives you two equations in two unknowns. As you have a quadratic, there are two solutions. They correspond to the two different circles that each pass through the given points. – Ross Millikan Apr 05 '17 at 18:23
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That is right. Now solve the two equations for $x,y$ You can subtract them to remove the squared terms from one, solve that for one of the variables, substitute into the other, and have a one variable quadratic. – Ross Millikan Apr 05 '17 at 18:42
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Let $\Omega $ be the desired circle. The equation of this circle is $\Omega : (x-a)^2 + (y-b)^2 = 5^2$ Where $I (a,b)$ is the center of $\Omega $ $$A \in \Omega\quad \text {and}\quad B \in \Omega\\ \begin {align} &\Leftrightarrow \begin {cases} (-8 - a)^2 + b^2 = 5^2\\ (-4 - a)^2 + (-2-b)^2 = 5^2 \end{cases} \\ &\Leftrightarrow \begin {cases} a^2 + 16a + 64 + b^2 - a^2 - 8a - 16 - b^2 - 4b - 4 = 0\\ a^2 + 8a + 16 + b^2 +4b +4 = 5^2 \end{cases} \\ &\Leftrightarrow \begin {cases} 8a - 4b + 44 = 0\\ a^2 + 8a + b^2 +4b -5 =0 \end{cases} \\ &\Leftrightarrow \begin {cases} 2a - b + 11 = 0\\ a^2 + 8a + b^2 +4b -5 =0 \end{cases} \\ &\Leftrightarrow \begin {cases} b = 2a + 11\\ a^2 + 8a + (2a+ 11)^2 + (44 +8a) -5 =0 \end{cases} \\ &\Leftrightarrow \begin {cases} b = 2a + 11\\ a^2 + 8a + 4a^2+ 44a + 121 +8a + 39 =0 \end{cases} \\ &\Leftrightarrow \begin {cases} b = 2a + 11\\ 5a^2+ 60a + 160 =0 \end{cases}\\ &\Leftrightarrow \begin {cases} b = 2a + 11\\ a^2+ 12a + 32 =0 \quad \Delta = 144 - 128 =16 \end{cases}\\ &\Leftrightarrow \begin {cases} b = 2a + 11\\ a_1= \frac {-12+4}2 \quad a_1= \frac {-12- 4}2 \end{cases}\\ &\Leftrightarrow \begin {cases} b = 2a + 11\\ a_1= -4 \quad a_2= -8 \end{cases}\end {align}\\ \Leftrightarrow (a_1,b_1) = (-4,3)\quad\text {or}\quad(a_2,8) = (-8,-5)$$ So there exists exaclty two circles:
The first $\Omega_1$ with center $(-4,3)$.
And the second $\Omega_2$ with center $(-8,-5)$.
Rami Zouari
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