The point of intersection of the direct common tangents drawn to the circles (x+11)2+(y-2)2=225 and (x-11)2+(y+2)2=25 is

Question

The point of intersection of the direct common tangents drawn to the circles (x+11)2+(y-2)2=225 and (x-11)2+(y+2)2=25 is (A) ((-11/2), 1) (B) (-22,4) (C) ((11/2),-1) (D) (22,-4)

Tardigrade Admin · Accepted Answer

The direct common tangents to two circles meet on the line of centres and divide it externally in the ratio of the radii centres of the two circles are

(- 11, 2)

and

(11, - 2)

and their radii are 15 and

5.

∴

Point of intersection

= (\frac{11 \times 15 - ( - 11 ) \times 5}{15 - 5}, \frac{- 2 \times 15 - 2 \times 5}{15 - 5})

= (\frac{165 + 55}{10}, \frac{- 30 - 10}{10}) = (22, - 4)

Q. The point of intersection of the direct common tangents drawn to the circles $(x + 11)^{2} + (y - 2)^{2} = 225$ and
$(x - 11)^{2} + (y + 2)^{2} = 25$ is

$(\frac{- 11}{2}, 1)$

$(- 22, 4)$

$(\frac{11}{2}, - 1)$

$(22, - 4)$

Q. The point of intersection of the direct common tangents drawn to the circles (x+11)2+(y−2)2=225 and (x−11)2+(y+2)2=25 is

(2−11​,1)

(−22,4)

(211​,−1)

(22,−4)

Q. The point of intersection of the direct common tangents drawn to the circles $(x + 11)^{2} + (y - 2)^{2} = 225$ and
$(x - 11)^{2} + (y + 2)^{2} = 25$ is

$(\frac{- 11}{2}, 1)$

$(- 22, 4)$

$(\frac{11}{2}, - 1)$

$(22, - 4)$