Question

Calc III... I've tried inputting (((x+1)^2)+((y-6)^2)+((z+5)^2))-86, and the input in the picture below,neither are correct.

Answer Preview Entered Result x² + 2x + y? + z² + 10z – 12y – 24 (x^2)+2*x+(y^2)+(z^2)+10*z-12*y-24 incorrect The answer above is NOT correct. (1 point) Find the equation of a sphere if one of its diameters has endpoints: (-6, 10, -7) and (4, 2, -3). x^2+2x+y^2+z^2+10z-12y-24 Note that you must move everything to the left hand side of the equation and that we desire the coefficients of the quadratic terms to be 1. HINT. The center of the sphere is the midpoint of its diameter. HINT. Let A(xA, YA, Z A) and B(xB,YB,- (хА + в) /2, Ус — 3, ZB) be endpoints of a line segment in the space. Then its midpoint C has coordinates (УА + ув)/2, 2с (ZA + ZB)/2. хс —

Answer 1

Soldes let are end A= (-6,10,-7) f B= ( 4,0,-3) point of diameters We now that midpoint formula x= - se on valoda 12=-73 x=-1, 7=6, 3 225 : ce 0=(-1,6,-5). are o is onto centre midpoint of diameters of the conte sphese find radius. po dro, A) = droiB) To JFB)?+62+ Crap2 4.01018)233 +624 22 8=52571674 dlo,B)= Jastitu 8=56945 dro,B) = 5 45 Radius of circle is 545 We know that is 021-16,-5) with equation spherectoate a radias 565 centre

(8+102 + (yng)2 + 12+5)² = (505) 202+20tl + y2-12y+36+ z2tlo ZtdS =45 202 + y2 +22+28-12y +102 +1750 This is required equation of spuese,