If the circles x 2 + y 2 - 16 x - 20 y + 164 = r 2   and ( x - 4 ) 2 + y - 7 2 = 36 intersect at two distinct points, then:

Question

If the circles x 2 + y 2 - 16 x - 20 y + 164 = r 2 &#8201; and ( x - 4 ) 2 + y - 7 2 = 36 intersect at two distinct points, then:

Aicharya · Accepted Answer

As we know, if two circles intersect each other, then

$|r_{1} - r_{2}| < C_{1} C_{2} < r_{1} + r_{2}$ ........(i)

Now for the first circle $C_{1} (8, 10)$ and $r_{1} = r$

For the second circle $C_{2} (4, 7)$ and $r_{2} = 6$

From (i)

$|r - 6| < 5 < r + 6$

$\Rightarrow |r - 6| < 5 . . . (ii)$ & $5 < r + 6 . . . (iii)$ $\{|r - 6| = \{\begin{cases} r - 6, r \geq 6 \\ 6 - r, r < 6 \end{cases}\}$

from $(iii)$ $r > - 1 . . . (iv)$

from $(ii)$

when $- 1 < r < 6$ then $6 - r < 5 \Rightarrow 1 < r . . . (v)$

when $r \geq 6 \Rightarrow r - 6 < 5 \Rightarrow r < 11 . . . (vi)$

from $(iv), (v), (vi)$ we get

$\Rightarrow r \in (1, 11)$

If the circles x 2 + y 2 - 16 x - 20 y + 164 = r 2   and ( x - 4 ) 2 + y - 7 2 = 36 intersect at two…

Solution

If the circles x 2 + y 2 - 16 x - 20 y + 164 = r 2 &#8201; and ( x - 4 ) 2 + y - 7 2 = 36 intersect at two…

Solution

If the circles x 2 + y 2 - 16 x - 20 y + 164 = r 2 and ( x - 4 ) 2 + y - 7 2 = 36 intersect at two…