Question

Two circles with diameters 68 cm and 40 cm, intersect each other and the length of their common chord is 32 cm. Find the distance between their centers

Answer 1

Step-by-step explanation:

1)If the two circles touch externally, then distance between their centres is = sum of their radius.

Therefore,

34+20 = 54

Therefore distance between their centres is 54 cm.

2) If the two circles touch internally, then distance between their centres is = difference of their radius.

Therefore,

34-20 =14

Therefore distance between their centres is 14 cm.

Answer 2

Given:

Two circles with center O and O' having diameter 68 cm and 40 cm respectively.

Length of the common chord = 32 cm

To find:

Distance between their centers.

Solution:

Step 1

Radius of circle with center O = 34 cm

Radius of circle with center O' = 20 cm

We have been given that the two circles intersect each other and they have a common chord (lets say AB) whose length is 32 cm.

Now,

If we see the attached diagram carefully,

A ΔBOA is formed whose perpendicular on AB (lets say at P) is the line joining the centers O and O'.

Step 2

Therefore,

In ΔOAP , Applying Pythagoras theorem, we get

$(OA)^{2} = (OP)^{2} + (AP)^{2}$ or

$(OP)^{2}=(OA)^{2}-(AP)^{2}$

Substituting the values, we get

$(OP)^{2}= (34)^{2}-(16)^{2}$ ;or

$(OP)^{2} = 900 \\OP = 30 cm$

Similarly a ΔO'AB is present

and in ΔO'AP , applying Pythagoras theorem, we get

$(O'A)^{2}=(O'P)^{2}+(AP)^{2} \\(O'P)^{2}= (O'A)^{2}-(AP)^{2}$

Substituting the values, we get

$(O'P)^{2} = (20)^{2}-(16)^{2} \\(O'P)^{2}=144\\O'P = 12cm$

Step 3

Distance between their centers $(00') = OP+O'P$

                                                             $= (30+12)cm$

  Distance between their centres      $= 42 cm$

Final answer:

Hence, the distance between the centers OO' will be 42 cm.

Attached image