MHT CET 2024 10th May Morning Shift | Circle Question 13 | Mathematics

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

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

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

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

MHT CET 2024 4th May Evening Shift

MCQ (Single Correct Answer)

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The equation of the circle which passes through the centre of the circle $x^2+y^2+8 x+10 y-7=0$ and concentric which the circle $2 x^2+2 y^2-8 x-12 y-9=0$ is

$x^2+y^2-4 x-6 y+77=0$

$x^2+y^2-4 x-6 y-89=0$

$x^2+y^2-4 x-6 y+97=0$

$x^2+y^2-4 x-6 y-87=0$

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