Question

The number of common tangents to
x2 + y2 = 256, (x - 3)+(y - 4)2 = 121 is
in one 2) two
3) four
4) zero​

Answer 1

The number of common tangents to  x² + y² = 256, (x - 3)²+(y - 4)² = 121 is one

Therefore, option (1) is correct.

Step-by-step explanation:

The given circles are

$x^2+y^2=256$

or $x^2+y^2=16^2$

The centre of this circle C1 is (0,0) and radius r1 is 16 units

And

$(x-3)^2+(y-4)^2=121$

$\implies (x-3)^2+(y-4)^2=11^2$

The centre of this circle C2 is (3, 2) and radius r2 11 units

Now,

Distance between the centre of the two circles

= C1C2

$=\sqrt{(3-0)^2+(4-0)^2)}$

$=\sqrt{25}$

$=5$

$=16-11$

$=r1-r2$

Therefore, the second circle is contained within the first circle and they touch each other at one point

Therefore, there will be only one common tangent for both the circles.

Hope this answer is helpful.

Know More:

Q: What are all the common tangents of the circle x^2 + y^2 = 9 and x^2 + y^2 - 16x +2y +49 = 0.

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