Question

The number of common tangents to the circles x2 + y2 + 2x + 8y 25 = 0 and x2 + y2 4x 10y + 19 = 0 is

Answer 1

Answer:

(In explanation)

Step-by-step explanation:

let's first look at $x^{2} + y^{2} + 2x + 8y + 25 =0$

It's Center is (-1,4) but its radius is $\sqrt{-8}$, which is not possible ( not going into imaginary numbers here)

Hence we can assume it to be a point circle.

So,now we have to find the equation of tangent of $x^{2} + y^{2} + 4x +10y +19 =0$ which passes through (-1,4)

the second circle can be reduced to: $(x+2)^{2}  + (y+5)^{2}= 10$

It's slope form equation is: $y+5 =m(x+2) + \sqrt{10m^{2}+10 }$

it passes through (-1,4)

so,$4 + 5 = m(-1 + 2) +\sqrt{10m^2 + 10}$

from here m can be found , and hence the common tangent

(if m doesn't have real values, it means that there is no common tangent between those circles)

Answer 2

The number of common tangents will be zero

Step-by-step explanation:

The given circles are

$x^2+y^2+2x+8y+25=0$

or $x^2+2x+1+y^2+2\times 4y+16-1-16-25=0$

or $(x+1)^2+(y+4)^2=42$

or $(x+1)^2+(y+4)^2=(\sqrt{42})^2$

The centre of this circle C1 is (-1,-4) and radius r1 is √42 units

And

$x^2+y^2+4x+10y+19=0$

$\implies x^2+2\times 2x+4+y^2+2\times 5y+25-4-25+19=0$

$\implies (x+2)^2+(y+5)^2=10$

$\implies (x+2)^2+(y+5)^2=(\sqrt{10})^2$

The centre of this circle C2 is (-2, -5) and radius r2 √10 units

Now,

Distance between the centre of the two circles

= C1C2

$=\sqrt{(-1+2)^2+(-4+5)^2)}$

$=\sqrt{2}$

$=1.414$ units

Difference of the radius of the circles

$=\sqrt{42}-\sqrt{10}$

$=6.48-3.16$

$=3.32$ units

Therefore, the second circle is contained within the first circle and they do not touch each other

Therefore, there will be no 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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