A circle is given by x^2 + y^2 - 6x + 8y - 11 | vedclass

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Question

(B) Let the equation of the circle be $S(x, y) = x^2 + y^2 - 6x + 8y - 11 = 0$. For point $P(0, 0)$: $S(0, 0) = 0^2 + 0^2 - 6(0) + 8(0) - 11 = -11$. S

Vedclass · Accepted Answer

One outside and one inside the circle

Vedclass · Answer

(B) Let the equation of the circle be $S(x, y) = x^2 + y^2 - 6x + 8y - 11 = 0$. For point $P(0, 0)$: $S(0, 0) = 0^2 + 0^2 - 6(0) + 8(0) - 11 = -11$. Since $S(0, 0) For point $Q(1, 8)$: $S(1, 8) = 1^2 + 8^2 - 6(1) + 8(8) - 11 = 1 + 64 - 6 + 64 - 11 = 112$. Since $S(1, 8) > 0$,the point $Q(1, 8)$ lies outside the circle. Therefore,one point lies inside and one point lies outside the circle.

$A$ circle is given by $x^2 + y^2 - 6x + 8y - 11 = 0$. Determine the position of the points $P(0, 0)$ and $Q(1, 8)$ with respect to the circle.

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