Question

$\theta =\frac{3\pi }{4}$ is a point on the circle $(x+2{)}^2+(y-3{)}^2=8$ . then the coordinates of the point are:

• A. (0,1)
• B. (-4,1)
• C. (0,5)
• D. (-4,5)

Answer 1

The correct option is D (-4,5)

$(x+2{)}^2+(y-3{)}^2=8$

radius $=2\sqrt{2}=r$

$\theta =\frac{3\pi }{4}$

coordinate of point $P$ will be

given by

$P(-2+r\cos \theta ,3+r\sin \theta )$

$\Rightarrow P(-2+2\sqrt{2}(-\frac{1}{\sqrt{2}}),3+2\sqrt{2}\times \frac{1}{\sqrt{2}})$

$\Rightarrow P(-2-2,3+2)$

$\Rightarrow P(-4,5)$