Circle ate circle

Geometry Level 4

If the circle x 2 + y 2 = 1 x^2+y^2=1 is completely contained in the circle x 2 + y 2 + 4 x + 3 y + k = 0 x^2+y^2+4x+3y+k=0 , then what is the largest integral value of k k ?


The answer is -7.

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1 solution

Rohit Ner
May 29, 2016

As the value of k k increases, the radius of the circle centered at ( 2 , 3 2 ) \left(-2,-\dfrac{3}{2}\right) decreases. When this circle touches the given circle which is centered at origin, the value of k k can be obtained as follows,

C D = A D + A C 2 2 + ( 3 2 ) 2 k = 1 + 2 2 + ( 3 2 ) 2 25 4 k = 7 2 k = 6 \begin{aligned} \overline { CD } &=\overline { AD } +\overline { AC }\\\sqrt{2^2+{\left(\dfrac{3}{2}\right)}^2-k}&=1+\sqrt{2^2+{\left(\dfrac{3}{2}\right)}^2}\\ \sqrt{\dfrac{25}{4}-k}&=\dfrac{7}{2}\\k&=-6\end{aligned}

This is the case when the circles touch each other internally.

Hence, as per the condition required the maximum integer value of k k is 7 \LARGE\color{#3D99F6}{\boxed{-7}} .

I entered -6, needed that extra step to earn the credit for that exercise.

Hana Wehbi - 5 years ago

The solution is perfect. The way I understood is:
The big circle should have a r adius of CA + AD= 1 + 2 2 + ( 3 2 ) 2 = 7 2 . 1+\sqrt{2^2+{\left(\frac{3}{2}\right)}^2}=\frac7 2.
So the equation of the big circle is:-
( x + 2 ) 2 + ( y + 3 2 ) 2 = ( 7 2 ) 2 . x 2 + 4 x + 4 + y 2 + 3 y + 9 4 = 49 4 . x 2 + y 2 + 4 x + 3 y 6 = 0. S o k = 6. (x+2)^2+(y+\frac 3 2)^2 =(\frac7 2 )^2.\\ \implies x^2+4x+4\ \ +\ \ y^2+3y+\frac 9 4 =\frac{49} 4.\\ \therefore\ \ x^2 + y^2 + 4x+3y - 6=0.\ \ \ \ So \ \ k=-6.


Niranjan Khanderia - 5 years ago

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