Question

Find the area of a circle whose centre is (7, 8) and which passes through the point (4, 10).

(Take π = 3.14)​

Answer 1

Step-by-step explanation:

the centre of circle co-ordinates is (7,8)

point through which the circle passes is (4,10)

radius=

$\sqrt{ {(diff. \: \: of \: abscissa)}^{2} - ( {diff. \: \: of \: ordinates)}^{2} }$

$= \sqrt{ {(7 - 4)}^{2} {(8 - 10)}^{2} }$

$= \sqrt{ {3}^{2} - {( - 2)}^{2} }$

$= \sqrt{9 -( 4) }$

$= \sqrt{5}$

therefore, area of circle=

$\pi {r}^{2}$

$= 3.14 \times \sqrt{5} \times \sqrt{5}$

$= 3.14 \times 5$

=$15.7 \: {units}^{2}$