AP EAPCET 2021 - 19th August Evening Shift | Circle Question 141 | Mathematics

If a foot of the normal from the point $$(4,3)$$ to a circle is $$(2,1)$$ and $$2 x-y-2=0$$, is a diameter of the circle, then the equation of circle is

$$x^2+y^2+2 x+1=0$$

$$x^2+y^2+2 x-1=0$$

$$x^2+y^2-2 x-1=0$$

$$2\left(x^2+y^2\right)-2 x-1=0$$

AP EAPCET 2021 - 19th August Evening Shift

MCQ (Single Correct Answer)

-0

The length of the tangent from any point on the circle $$(x-3)^2+(y+2)^2=5 r^2$$ to the circle $$(x-3)^2+(y+2)^2=r^2$$ is 16 units, then the area between the two circles in square units is

$$32 \pi$$

$$4 \pi$$

$$8 \pi$$

$$256 \pi$$

AP EAPCET 2021 - 19th August Evening Shift

MCQ (Single Correct Answer)

-0

The equation of the circle, which cuts orthogonally each of the three circles

$$\begin{aligned} & x^2+y^2-2 x+3 y-7=0, \\ & x^2+y^2+5 x-5 y+9=0 \text { and } \\ & x^2+y^2+7 x-9 y+29=0 \end{aligned}$$

$$x^2+y^2-16 x-18 y-4=0$$

$$x^2+y^2=a^2$$

$$x^2+y^2-16 x=0$$

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

AP EAPCET 2021 - 19th August Morning Shift

MCQ (Single Correct Answer)

-0

Find the equations of the tangents drawn to the circle $$x^2+y^2=50$$ at the points where the line $$x+7=0$$ meets it.

$$7 x+y+50=0$$ and $$7 x-y+50=0$$

$$x+y=0$$ and $$x-y=0$$

$$x+7 y+5=0$$ and $$y-7 x+5=0$$

$$x+7 y+50=0$$ and $$x-7 y+50=0$$

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