A(4,3) and B(2,5) are two points. If P is a v | vedclass

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(B) $1$. Find the midpoint $M$ of $AB$: $M = (\frac{4+2}{2}, \frac{3+5}{2}) = (3, 4)$.$2$. The distance of $P(x, y)$ from $M(3, 4)$ is at most $5$ uni

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$x^2+y^2-6x-8y \leq 0, x+y-7 < 0$

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(B) $1$. Find the midpoint $M$ of $AB$: $M = (\frac{4+2}{2}, \frac{3+5}{2}) = (3, 4)$.$2$. The distance of $P(x, y)$ from $M(3, 4)$ is at most $5$ units: $(x-3)^2 + (y-4)^2 \leq 5^2$,which simplifies to $x^2 + y^2 - 6x - 8y \leq 0$.$3$. The equation of line $AB$ is $y - 3 = \frac{5-3}{2-4}(x - 4)$ $\Rightarrow y - 3 = -1(x - 4)$ $\Rightarrow x + y - 7 = 0$.$4$. The origin $(0, 0)$ satisfies $0 + 0 - 7 = -7 $5$. Combining these,the locus is $x^2 + y^2 - 6x - 8y \leq 0$ and $x + y - 7 < 0$.

$A(4,3)$ and $B(2,5)$ are two points. If $P$ is a variable point on the same side as the origin with respect to the line $AB$ and is at most at a distance of $5$ units from the midpoint of $AB$,then the locus of $P$ is

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