Find the equation of the circle orthogonal to the circles $x^{2} + y^{2} + 3 x - 5 y + 6 = 0 a n d 4 x^{2} + 4 y 2 - 28 x + 29 = 0$ and whose center lies on the line 3x + 4y + 1 = 0.

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Find the maximum value of y which satisfy the inequations y2+5y+4≤0 and y2−2y−15≥0

The correct option is A −3y2+5y+4≤0y2+4y+y+4≤0(y+1)(y+4)≤0y∈(−4,−1)&y2−2y−15≥0y2−5y+3y−15≥0(y+3)(y−5)≥0y∈(−∞,−3)∪(5,∞)∴ Maximum value of y=−3

Find the maximum value of y which satisfy the inequations $y^{2} + 5 y + 4 \leq 0$ and $y^{2} - 2 y - 15 \geq 0$

The correct option is A $- 3$
$y^{2} + 5 y + 4 \leq 0 y^{2} + 4 y + y + 4 \leq 0 (y + 1) (y + 4) \leq 0 y \in (- 4, - 1) & y^{2} - 2 y - 15 \geq 0 y^{2} - 5 y + 3 y - 15 \geq 0 (y + 3) (y - 5) \geq 0 y \in (- \infty, - 3) \cup (5, \infty) ∴$
Maximum value of $y = - 3$