Circle

Given, lines arex+2y−2=0and
2x+3y−1=0
Solving these two lines, we get
x=−4‌ and ‌y=3
Thus, the center of circle is(−4,3).
Since, the general eq. of circle is(x−h)2+(y−k)2=r2, where(h,k)is the center andris the radius.
So,(x−(−4))2+(y−3)2=r2
⇒‌‌(x+4)2+(y−3)2=r2‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
But, the circle passes through the point(8,8),
So,(8+4)2+(8−3)2=r2
122+52=r2
169=r2
⇒‌‌r=13
Now, from Eq. (i), we get
(x+4)2+(y−3)2=r2‌=169
⇒‌‌x2+8x+16+y2−6y+9‌=169
⇒‌‌x2+y2+8x−6y−144‌=0
Comparing with the given equation
x2+y2+px+qx+r=0
We getp=8,q=−6,r=−144
∴‌‌p2+q2+r‌=82+(−6)2+(−144)
‌=64+36−144
‌=−44