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

Question

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

Accepted Answer

Circles are x2 + y2 - 16x - 20y + 164 = r2 $$ \Rightarrow $$ c1 (8, 10)

and (x - 4)2 + (y - 7)2 = 36

they intersect at two distinct points

$$\left| {{r_1} - {r_2}} \right| < {c_1}{c_2} < {r_1} + {r_2}\left\{ {{c_1}{c_2} = \sqrt {16 + 9}  = 5} \right\}$$

Now  $$\left| {r - 6} \right| < 5 < r + 6$$

$$\left| {r - 6} \right| < 5$$

$$ \Rightarrow $$  $- 5 < r - 6 < 5$

$$ \Rightarrow $$  $$1 < r < 11\,\,\,\,\,\,\,\,\,...(i)$$

$5 < r + 6$

$$ - 1 < r\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$$

from (i) and (ii)

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

Solutions · 1