The point (p, q) lies in the circle x^2 + y^2 = 289 and at the line y = 15. The value of P that satisfies is?

1 Answer
Mar 3, 2018

#p=+-8#

Explanation:

As #(p,q)# lies on the line #y=15#, its ordinate is #15#, while abscissa given by #p# is not known. As #p# lies on #x^2+y^2=289#

we have #p^2+225=289#

or #p^2)=289-225=64#

and #p=+-8#

Hence, points are #(8,15)# and #(-8,15)#