If the lines x + y = 6 and x + 2y = 4 are dia | vedclass

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(A) Given that the diameter of the circle is $20$,the radius $r = \frac{20}{2} = 10$.The center of the circle is the point of intersection of the two

Vedclass · Accepted Answer

$x^2 + y^2 - 16x + 4y - 32 = 0$

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(A) Given that the diameter of the circle is $20$,the radius $r = \frac{20}{2} = 10$.The center of the circle is the point of intersection of the two diameters $x + y = 6$ and $x + 2y = 4$.Subtracting the first equation from the second: $(x + 2y) - (x + y) = 4 - 6 \Rightarrow y = -2$.Substituting $y = -2$ into $x + y = 6$,we get $x - 2 = 6 \Rightarrow x = 8$.So,the center of the circle is $(h, k) = (8, -2)$.The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$.$(x - 8)^2 + (y + 2)^2 = 10^2$.$x^2 - 16x + 64 + y^2 + 4y + 4 = 100$.$x^2 + y^2 - 16x + 4y + 68 - 100 = 0$.$x^2 + y^2 - 16x + 4y - 32 = 0$.

If the lines $x + y = 6$ and $x + 2y = 4$ are diameters of a circle whose diameter is $20$,then the equation of the circle is:

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