Question

The number of common tangents to the two circles x2+y2−8x+2y=0x2+y2−8x+2y=0  and x2+y2−2x−16y+25=0
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Answer 1

The number of common tangents will be two

Step-by-step explanation:

The first circle (C1) is

$x^2+y^2-8x+2y=0$

or $x^2-2\times 4\times x+16+y^2+2\times y+1-16-1=0$

or $(x-4)^2+(y+1)^2=17$

or $(x-4)^2+(y+1)^2=(\sqrt{17})^2$

The centre of this circle C1 is (4,-1) and radius r1 is √17 units

And, the second circle (C2) is

$x^2+y^2-2x-16y+25=0$

$\implies x^2-2\times x+1+y^2-2\times 8y+64-1-64+25=0$

$\implies (x-1)^2+(y-8)^2=40$

$\implies (x-1)^2+(y-8)^2=(\sqrt{40})^2$

The centre of this circle C2 is (1, 8) and radius r2 √40 units

Now,

Distance between the centre of the two circles

= C1C2

$=\sqrt{(4-1)^2+(-1-8)^2)}$

$=\sqrt{9+81}$

$=\sqrt{90}$

$=9.49$ units

Sum of the radius of the circles

$=\sqrt{40}+\sqrt{17}$

$=6.32+4.12$

$=10.44$ units

Since the distance between the centre of the circles is less than the sum of their radius

Therefore, the circles intersect each other

Therefore, the no. of common tangent for both the circles will be two

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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