Question

Two circles of radii 10 cm and 8 cm intersect at two points and the length of the common chord is 12 cm. Find the distance between their centres.

Answer 1

Proof: In circle 'a' OP is radius. PQ is chord, $OX\perp PQ$
$\Rightarrow PX=\frac{1}{2}PQ\to \perp$ from centre to chord bisects chord
PQ=12 cm
$\Rightarrow PX=6cm$
$PO-PX=\frac{1}{2}OX$
$\Rightarrow 10-6=4cm$
$\therefore OX=8cm$
Although, you are getting the right ans, here but the method is wrong
OE - OX = XE
10 - 8 cm = 2 cm
$\therefore$ XE = 2 cm

2) In circle 'b'
OP is radius
PQ is chord
$O^{{}^{\prime }}X\perp PQ\Rightarrow PX=\frac{1}{2}PO\to proved\Rightarrow PX=6cm\to provedP{O}^{{}^{\prime }}-PX=\frac{1}{2}{O}^{{}^{\prime }}X$
8 - 6 = 2 cm
$\therefore O^{{}^{\prime }}X=4cm{O}^{{}^{\prime }}D-O^{{}^{\prime }}X=XD8-4=4cm\therefore XD=4cm$

3) $\therefore O{O}^{{}^{\prime }}=(OX+O^{{}^{\prime }}X)-(DX+XE)$
=(8+4)-(2+4)
=12-6
=6 cm

1) We cannot say that,OX = 8 cm
Consider $\Delta OPX$
OP is radius

OX has to be calculated using pythagoras theorem.
${10}^2=6^2+O{X}^2$
$\Rightarrow O{X}^2=100-36=64$
$\Rightarrow OX=8cm$

2) $|||ly$ in $\Delta P{O}^{{}^{\prime }}X$
$P{O}^{{}^{\prime 2}}=P{X}^2+X{O}^{{}^{\prime 2}}\Rightarrow 8^2=6^2+X{O}^{{}^{\prime 2}}\Rightarrow X{O}^{{}^{\prime 2}}=64-36=28\Rightarrow X{O}^{{}^{\prime }}=\sqrt{28}$
$\therefore$ Distance between centres