MHT CET 2024 10th May Evening Shift | Circle Question 12 | Mathematics

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)

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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

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

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

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

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

MHT CET 2024 9th May Evening Shift

MCQ (Single Correct Answer)

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The equation of the circle, concentric with the circle $x^2+y^2-6 x-4 y-12=0$ and touching the $\mathrm{X}$-axis is

$x^2+y^2-6 x-4 y+5=0$

$x^2+y^2-6 x-4 y+17=0$

$x^2+y^2-6 x-4 y+9=0$

$x^2+y^2-6 x-4 y+4=0$

MHT CET 2024 9th May Morning Shift

MCQ (Single Correct Answer)

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The equation of the circle, the end points of whose diameter are the centres of the circles $x^2+y^2+6 x-14 y+5=0$ and $x^2+y^2-4 x+10 y-4=0$ is

$x^2+y^2-x-2 y+41=0$

$x^2+y^2+x-2 y-41=0$

$x^2+y^2-x+2 y-41=0$

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