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
