Question

If (4, 3) and (-12, -1) are end points of a
diameter of a circle, then the equation of the
circle is-​

Answer 1

Answer:

this can't be possible as diameter can't be in negative nom

Answer 2

The equation of the circle with ends of diameter (4, 3) and (-12, -1) is $x^{2}+y^{2}+8x-2y-51=0$

• The equation of circle with ends of the diameter as $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is given by

                $(x-x_{1})(x-x_{2})+(y-y_{1})(y-y_{2})=0-----(1)$

• Given, ends of the diameter are

                $(x_{1},y_{1})=(4,3)$

                 $(x_{2},y_{2})=(-12,-1)$

• Substituting above values in (1), we get

                $(x-4)(x+12)+(y-3)(y+1)=0$

                 $x^{2}+12x-4x-48+y^{2}+y-3y-3=0$

                 $x^{2}+y^{2}+8x-2y-51=0$

• Therefore, the equation of required circle is $x^{2}+y^{2}+8x-2y-51=0$