JEE MAIN - Mathematics (2020 - 9th January Evening Slot - No. 18)
If the curves, x2 – 6x + y2 + 8 = 0 and
x2 – 8y + y2 + 16 – k = 0, (k > 0) touch each other at a point, then the largest value of k is ______.
x2 – 8y + y2 + 16 – k = 0, (k > 0) touch each other at a point, then the largest value of k is ______.
Answer
36
Explanation
C1 : x2 + y2 – 6x + + 8 = 0
C1(3, 0) and r1 = 1
C2 : x2 + y2 – 8y + 16 – k = 0
C2(0, 4) and r2 = $$\sqrt k $$
Two circles touch each other
$$ \therefore $$ C1C2 = | r1 $$ \pm $$ r2 |
$$ \Rightarrow $$ 5 = | 1 $$ \pm $$ $$\sqrt k $$ |
$$ \therefore $$ 1 + $$\sqrt k $$ = 5 or $$\sqrt k $$ - 1 = 5
$$ \Rightarrow $$ k = 16 or k = 36
So largest value of k = 36.
C1(3, 0) and r1 = 1
C2 : x2 + y2 – 8y + 16 – k = 0
C2(0, 4) and r2 = $$\sqrt k $$
Two circles touch each other
$$ \therefore $$ C1C2 = | r1 $$ \pm $$ r2 |
$$ \Rightarrow $$ 5 = | 1 $$ \pm $$ $$\sqrt k $$ |
$$ \therefore $$ 1 + $$\sqrt k $$ = 5 or $$\sqrt k $$ - 1 = 5
$$ \Rightarrow $$ k = 16 or k = 36
So largest value of k = 36.
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