Question

The number of common tangents to the circles
x² + y² - 2x - 4y+1=0
x2 + y2-12x-16y +91 = 0 is
(a) 1
(c)3
(d) 4
(b)2
PLZ ANSWER WITH FULL STEPS. SHORTCUT ANSWER WILL BE REPORTED​

Answer 1

Answer:

Number of common tangents are 4

Step-by-step explanation:

follow the above steps

Answer 2

The number of common tangents will be 4

Therefore, option (d) is correct.

Step-by-step explanation:

The given circles are

$x^2+y^2-2x-4y+1=0$

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

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

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

The centre of this circle C1 is (1,2) and radius r1 is 2 units

And

$x^2+y^2-12x-16y+91=0$

$\implies x^2-2\times 6x+36+y^2-2\times 8y+64-36-64+91=0$

$\implies (x-6)^2+(y-8)^2=9$

$\implies (x-6)^2+(y-8)^2=3^2$

The centre of this circle C2 is (6, 8) and radius r2 3 units

Now,

Distance between the centre of the two circles

= C1C2

$=\sqrt{(6-1)^2+(8-2)^2)}$

$=\sqrt{25+36}$

$=\sqrt{61}$

$=7.81$ units

Sum of the radius of the circles

$=3+2$

$=5$ units

Since, the distance between the centre of the circles is greater than the sum of their radii

Therefore, both the circles do not touch or intersect each other

Thus, there will be 4 commom tangents .

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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