Question

The diameter of the bigger circle is 10cm and the radius of the smaller circle is 4cm. If the length of the common tangent is 40cm then the distance between their centres is? ​

Answer 1

Please refer above's diagram !!

In above's diagram

$C_1$ = Circle with the radius of 5 cm or diameter of 10 cm

$C_2$ = Circle with radius of 4 cm

Let AB be the common tangent

Given that length of AB = 40 cm

Now

Let H$C_2$ line parallel to AB

And H lie on the line A$C_1$

Then

Length of H$C_2$ = 40 cm as AH$C_2$B forms a parallelogram

Then

Now in ∆ H$C_2 C_1$

H$C_2$ = 40 cm

H$C_1$ = 1

( because as AH = B$C_2$ = 4 cm => H$C_1$ = 5 - 4 = 1 cm )

Now from Pythagorean theorem

which states that

$h^2 = p^2 + b^2$

as

h (hypotenuse ) = $C_1C_2$ = x cm

p (perpendicular) = H$C_2$ = 40 cm

b (base) = H$C_1$ = 1 cm

Then by putting value

=> $x^2 = (40)^2 + 1^2$

=> $x^2 = 1600 + 1$

=> $x^2 = 1601$

=> x = $\sqrt{1601}$

or x = 40.012498

or x = 40. 0125

Hope it was helpful

Answer 2

PU = radius of bigger = 5

CS = radius of smaller = 4

LM= Length of common tangent = 40

IN CLO and PMO

<L = <M = 90

<COL = POM

THUS AAA

SO CO/OP= OL/OM = CL/PM


CL /PM = 4/5


OM= OL + LM = OL + 40

4/5 = OL/( OL +40)

4( OL + 40) = 5 OL

4 OL+ 160 =5 OL

OL = 160

IN OLC

OC^2= OL^2 + LC^2

= 160^2 + 16

= 25600 +16 = 25616

Take OC also 160 as 160^2 = 25600 approx

OC/OP = 4/5

160/OP = 4/5

4 OP= 800

OP = 200

CP = 200 - 160 = 40

So distance between centers is 40