Two circles of radii 10 cm and 8 cm intersect at the points P and Q. If PQ = 12 cm, and the distance between the centers of the circles is x cm. The value of x (correct to one decimal place) is:

As per the given question,

Two circles are getting intersect at the point P and Q, as per the given below,

$$r_1=10cm $$

and $$r_2=8cm$$

$$\Rightarrow PR=RQ=\dfrac{PQ}{2}=6cm$$

We know that O'R and OR will be the perpendicular to the chord PQ and OO' have the length x.

Now, In $$\triangle RO'P$$ and $$\triangle PRO$$

$$\Rightarrow OO'^2=PR^2+RO'^2$$ and $$PO^2 = PR^2+RO^2$$

Now substituting the values in the above,

$$\Rightarrow 10^2=6^2+RO'^2$$ and $$8^2 = 6^2+RO^2$$

$$\Rightarrow RO'^2=100-36$$ and $$RO^2=64-36$$

$$\Rightarrow RO'^2=64$$ and $$RO^2=28$$

$$\Rightarrow RO'=8$$ and $$RO=2\sqrt{7}$$

Hence $$x = RO'+RO=8+2\sqrt{7}=13.3cm$$