MHT CET 2024 11th May Evening Shift | Circle Question 5 | Mathematics

$x=\frac{\mathrm{a}}{2}+\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{2} \cos \theta, y=\frac{\mathrm{b}}{2}+\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{2} \sin \theta$

$x=\frac{-\mathrm{a}}{2}+\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{4} \sin \theta, y=\frac{-\mathrm{b}}{2}+\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{4} \cos \theta$

$x=\frac{\mathrm{a}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{2}} \sin \theta, y=\frac{\mathrm{b}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{2}} \cos \theta$

$x=\frac{\mathrm{a}}{2}+\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{4} \cos \theta, y=\frac{\mathrm{b}}{2}+\frac{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}{4} \sin \theta$

MHT CET 2024 11th May Morning Shift

MCQ (Single Correct Answer)

-0

The equation of the tangent to the circle, given by $x=5 \cos \theta, y=5 \sin \theta$ at the point $\theta=\frac{\pi}{3}$ on it , is

$x-\sqrt{3} y=-5$

$x+\sqrt{3} y=10$

$\sqrt{3} x+y=5 \sqrt{3}$

$\sqrt{3} x-y=0$

MHT CET 2024 10th May Evening Shift

MCQ (Single Correct Answer)

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Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius $r$. If PS and RQ intersect at a point X on the circumference of the circle, then 2 r equals

$\sqrt{\mathrm{PQ} \cdot \mathrm{RS}}$

$\frac{\mathrm{PQ}+\mathrm{RS}}{2}$

$\frac{2 \cdot P Q \cdot R S}{P Q+R S}$

$\sqrt{\frac{\mathrm{PQ}^2+\mathrm{RS}^2}{2}}$

MHT CET 2024 10th May Morning Shift

MCQ (Single Correct Answer)

-0

The abscissae of the two points A and B are the roots of the equation $x^2+2 a x-b^2=0$ and their ordinates are roots of the equation $y^2+2 p y-q^2=0$. Then the equation of the circle with AB as diameter is given by

$x^2+y^2-2 \mathrm{a} x-2 \mathrm{p} y+\left(\mathrm{b}^2+\mathrm{q}^2\right)=0$

$x^2+y^2-2 \mathrm{a} x-2 \mathrm{p} y-\left(\mathrm{b}^2+\mathrm{q}^2\right)=0$

$x^2+y^2+2 \mathrm{a} x+2 \mathrm{p} y+\left(\mathrm{b}^2+\mathrm{q}^2\right)=0$

$x^2+y^2+2 \mathrm{a} x+2 \mathrm{p} y-\left(\mathrm{b}^2+\mathrm{q}^2\right)=0$

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