Find the equation of a circle with radius 10 and tangent to the line 3x + 2y = 5 at the point (1,1)
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what have you tried? – Lion Heart Apr 18 '22 at 12:21
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r² = (x-h)² + (y-k)² – Habolaji samuel Apr 18 '22 at 12:23
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1That is a first step. Add that $r=10$, that $(1, 1)$ is on the circle and that $(h-1, k-1)$ is orthogonal to the line $3 x + 2 y = 5$. – Gribouillis Apr 18 '22 at 12:29
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Since the line $3x+2y=5$ is tangent to the circle at $(1,1)$, the radius connecting the point $(1,1)$ and the centre $(a,b)$ has slope $\frac{2}{3}$. Therefore, the line segment decreases in $y$ by $2z$ and in $x$ by $3z$ when passing from $(1,1)$ to $(a,b)$ for some real number $z$. Further, since the radius is $10$, $z=\frac{10}{\sqrt{13}}$ by Pythagoras's Theorem. Therefore, $(a,b)=(1-\frac{30}{\sqrt{13}}, 1-\frac{20}{\sqrt{13}})$. Therefore, the equation of this circle is $(x-1+\frac{30}{\sqrt{13}})^2+(y-1+\frac{20}{\sqrt{13}})^2=100$.
George Adams
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The wording of the answer suggests that the solution is unique, which is not the case. – Théophile Apr 18 '22 at 14:42
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Let the coordinates of center $C(h;k)$ Since $r=10$, distance between center of the circle and point $(1;1)$ $(h-1)^2+(k-1)^2=10^2$
$m_{tangent}=-\frac{3}{2}$
$m_{normal}=\frac{2}{3}$
$m_{normal}=\frac{2}{3}=\frac{k-1}{h-1}$
$k-1=\frac{2}{3}(h-1)$
$(h-1)^2+\frac{4}{9}(h-1)^2=10^2$
$h=1-\frac{30}{\sqrt{13}},k=1-\frac{20}{\sqrt{13}}$
$h=1+\frac{30}{\sqrt{13}},k=1+\frac{20}{\sqrt{13}}$
equation of the circle(s) is $(x-h)^2+(y-k)^2=10^2$
Lion Heart
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