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i was given $$(X-1)^2+(Y+1)^2 =5$$, it said when $k$ maximum, this line $X+2Y-k=0$ will tangent the circle.

i was asked to find $k$.

Previous section we denoted $k$ for $\log_2 xy^2$ and $X=\log_2 x; Y=\log_2 y$

I used $$y=mx + r\sqrt{m^2+1}$$ and i get $2Y + X - 5 $. But the answer is $k=4$, what should i do?

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Dini
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2 Answers2

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You can get family of tangents: $$y+1=m(x-1)\pm \sqrt{5} \sqrt{1+m^2}~~~~(1),$$ where $m$ is a real number. So if $x+2y-k=0$ touches the given circle then perpendicular on it from the center $(1,-1)$ should equal the radius: $$|\frac{1-2-k}{\sqrt{5}}|=\sqrt{5} \implies (1+k)^2=25 \implies k=4,-6.$$ One may also use (1) to get to this answer by re-writing $x+2y-k$ as $$x-1+2(y+1)-1-k \implies (y+1)=-\frac{1}{2}(x-1)+(1+k)/2$$ This means $m=-1/2$ so the intercept $$(1+k)/2 =\pm \sqrt{5} \sqrt{1+m^2}$$

Z Ahmed
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  • Yes. Thanks. Accept answer in 10 minutes – Dini Feb 20 '20 at 14:42
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    @Lifeforbetter, See https://math.stackexchange.com/questions/774250/finding-the-equations-of-the-lines-and-tangent-to-the-circle and https://math.stackexchange.com/questions/2254073/the-line-y-mxc-is-a-tangent-to-x2y2-a2-if – lab bhattacharjee Feb 20 '20 at 14:51
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Use the fact the distance of any tangent from the center is equal to the radius of the circle

So, $$\sqrt5=\dfrac{|1+2-k|}{\sqrt{1^2+2^2}}$$

lab bhattacharjee
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