Question

Two circles of redii 10 and 8 intersect each other and lenght of common chord is 12 .Find distance between there centre. ans.is8+2√7 how?​

Answer 1

Step-by-step explanation:

Let P and Q be the centres of the circles which are intersecting each other at points A and B. Draw segment AB which is common chord of both circles.

Darw radii PA, PB & QA, QB.

Draw segment PQ which intersects AB at M.

$\therefore AM= BM= \frac{1}{2} \times AB\\(\because PM \:\& \:QM \: are \:perpendicular\\bisectors \: of \: AB) \\\therefore AM= BM= \frac{1}{2} \times 12\\\therefore AM= BM= 6$

Let PA = PB = 10 & QA = QB = 8

$In\: right\:\triangle\: PMA, \:PA \: is\:hypotenuse. \\</p><p>\therefore PM^2 = PA^2 - AM^2\\</p><p> \therefore PM^2 = 10^2 - 6^2</p><p>\\\therefore PM^2 = 100 - 36</p><p>\\\therefore PM^2 = 64</p><p>\\\therefore PM = \sqrt {64} </p><p>\\\huge\fbox {\therefore PM = 8}$

$In\: right\:\triangle\: QMA, \:QA \: is\:hypotenuse. \\</p><p>\therefore QM^2 = QA^2 - AM^2\\</p><p> \therefore QM^2 = 8^2 - 6^2</p><p>\\\therefore QM^2 = 64 - 36</p><p>\\\therefore QM^2 = 28</p><p>\\\therefore QM = \sqrt {28} </p><p>\\\huge\fbox {\therefore QM = 2\sqrt {7} }$

Now,

$PQ = PM + QM \\</p><p>\huge\fbox {\therefore PQ = 8 +2\sqrt {7}}$

Hence, the distance between centre of the circles is$(8 +2\sqrt {7}) \: units.$