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The area of circle centred at (1, 2) and passing through (4, 6) is

$5 π$
$10 π$
$25 π$
None of these

hint Hint:

A circle closed plane geometric shape. In technical terms, a circle is a locus of a point moving around a fixed point at a fixed distance away from the point.

The correct answer is: $25 pi$

Here, we have to find the area of circle.
Now, the centre of circle is C (1, 2).
The circle passes through the point P(4, 6).
Radius = CP
= √(4-1) ² + (6-2) ²
= √(3) ² + (4) ²
= √9+16
=√25
=5
Therefore, radius is 5 unit.
Now, Area of circle = π × 5 ×5
= 25π .
Hence, the correct option is (a).

The area of a circle is the space occupied by the circle in a two-dimensional plane. Area of circle = πr² (π = 22/7).

Three rods of same dimensions are arranged as shown in figure. They have thermal conductivities $K subscript 1 end subscript comma K subscript 2 end subscript$ and $K subscript 3 end subscript$ . The points $P$ and $Q$ are maintained at different temperatures for the heat to flow at the same rate along $P R Q$ and $P Q$ then which of the following options is correct

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Lux is equal to

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A 60 watt bulb is hung over the center of a table $4 m cross times 4 m$ at a height of 3 m. The ratio of the intensities of illumination at a point on the centre of the edge and on the corner of the table is

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If f(x) is a polynomial of degree five with leading coefficient one such that f(1)=1²,f(2)=2²,f(3)=3²,f(4)=4²,f(5)=5² then

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Maximum kinetic energy (E_k ) of a photoelectron varies with the frequency ( v ) of the incident radiation as

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The degree of the differential equation satisfying $square root of 1 plus x squared end root plus square root of 1 plus y squared end root equals A open parentheses x square root of 1 plus y squared end root minus y square root of 1 plus x squared end root close parentheses$ is

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A simple magnifying lens is used in such a way that an image is formed at 25 cm away from the eye. In order to have 10 times magnification, the focal length of the lens should be

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The area of circle centred at (1, 2) and passing through (4, 6) is

None of these

A circle closed plane geometric shape. In technical terms, a circle is a locus of a point moving around a fixed point at a fixed distance away from the point.

The correct answer is:

Here, we have to find the area of circle.Now, the centre of circle is C (1, 2).The circle passes through the point P(4, 6).Radius = CP= √(4-1) 2 + (6-2) 2= √(3) 2 + (4) 2= √9+16=√25=5Therefore, radius is 5 unit.Now, Area of circle = π × 5 ×5= 25π .Hence, the correct option is (a).

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$5 π$
$10 π$
$25 π$
None of these

The correct answer is: $25 pi$

In the following diagram if $V subscript 2 end subscript greater than V subscript 1 end subscript$ then

In the following diagram if $V subscript 2 end subscript greater than V subscript 1 end subscript$ then

The maximum kinetic energy (E_k ) of emitted photoelectrons against frequency v of incident radiation is plotted as shown in fig The slope of the graph is equal to

The maximum kinetic energy (E_k ) of emitted photoelectrons against frequency v of incident radiation is plotted as shown in fig The slope of the graph is equal to

In centre of the triangle formed by the lines y = x, y = 3x and y = 8 – 3x is

In centre of the triangle formed by the lines y = x, y = 3x and y = 8 – 3x is

A 60 watt bulb is hung over the center of a table $4 m cross times 4 m$ at a height of 3 m. The ratio of the intensities of illumination at a point on the centre of the edge and on the corner of the table is

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If f(x) is a polynomial of degree five with leading coefficient one such that f(1)=1²,f(2)=2²,f(3)=3²,f(4)=4²,f(5)=5² then

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Maximum kinetic energy (E_k ) of a photoelectron varies with the frequency ( v ) of the incident radiation as

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The degree of the differential equation satisfying $square root of 1 plus x squared end root plus square root of 1 plus y squared end root equals A open parentheses x square root of 1 plus y squared end root minus y square root of 1 plus x squared end root close parentheses$ is

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