Question

if 2 diameters of a circle lie along the lines x-y=9 and x-2y=7 and the area of the circle is 38.5sq cm find the equation if cire

Answer 1

Answer:

The equation of the required circle is

$\boxed{\bf\,4x^2+4y^2-88x-16y-451=0}$

Step-by-step explanation:

We know that the centre of the circle is the point of intersection of diameters.

x-y=9.........(1)

x-2y=7......(2)

(2)-(1) gives, y=2

put y=2 in (1), we get

x-2=9

x=9+2

x=11

$\therefore$centre C is (11,2)

Also,

Area of the circle = 38.5 square cm

$\pi\,r^2=38.5$

$\frac{22}{7}\,r^2=38.5$

$\frac{2}{7}\,r^2=3.5$

$\implies\,r^2=\frac{3.5*7}{2}$

$\implies\,r^2=0.5*0.5*7*7$

$\implies\,r=0.5*7$

$\implies\,r=3.5$

$\implies\,r=\frac{7}{2}\:cm$

The equation of the required circle is

$\boxed{\bf\,(x-h)^2+(y-k)^2=r^2}$

$\implies\;(x-11)^2+(y-2)^2=(\frac{7}{2})^2$

$\implies\;(x-11)^2+(y-2)^2=\frac{49}{4}$

$\implies\;x^2+121-22x+y^2+4-4y=\frac{49}{4}$

$\implies\;4x^2+484-88x+4y^2+16-16y=49$

$\implies\boxed{\bf\,4x^2+4y^2-88x-16y-451=0}$

Answer 2

Hope! it's helpful for you☺️☺️☺️