Question 989017
.
(a)
.
Standard form for circle:
.
{{{(x-h)^2+(y-k)^2=r^2}}} 
.
We know k=0, since center is on x axis. Using point (-3,1):
.
{{{(-3-h)^2+(1-0)^2=r^2}}}
{{{9+6h+h^2+1=r^2}}}
{{{h^2+6h+10=r^2}}}
.
Using point (-1,5)
{{{(-1-h)^2+(5-0)^2=r^2
{{{1+2h+h^2+25=r^2}}}
{{{h^2+2h+26=r^2}}}
.
Since both =r^2:
.
{{{h^2+6h+10=h^2+2h+26}}}
{{{4h-16=0}}}
{{{4h=16}}}
{{{h=4}}}
.
ANSWER (a): Co-ordinates of the center: (h,k)=(4,0)
.
(b) Find the equation:
{{{(x-h)^2+(y-k)^2=r^2}}} (h,k)=(4,0)
{{{(x-4)^2+(y-0)^2=r^2}}} Use point (-3,1) to find r^2
{{{(-3-4)^2+(1-0)^2=r^2}}}
{{{(-7)^2+(1)^2=r^2}}}
{{{49+1=r^2}}}
{{{50=r^2}}}
.
ANSWER:(b): Equation of circle: {{{(x-4)^2+y^2=50}}}
.
(c) Point (1,p)
Using value (1)for x (p=y):
{{{(1-4)^2+y^2=50}}}
{{{(-3)^2+y^2=50}}}
{{{9+y^2=50}}}
{{{y^2=41}}}
{{{y}}}=(+ or -){{{sqrt(41)}}}
.
ANSWER: Possible values of p are (+ or -){{{sqrt(41)}}}
.