Question

find the equation of the tangent at (1,1) to the circle 2x^2+2y^2-2x-5y+3=0
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Answer 1

Answer:

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The equation of the normal of the circle 2x

2

+2y

2

−2x−5y−7=0 passing through the point (1,1) is

Medium

Solution

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Correct option is

A

x+2y−3=0

Centre of the given circle is (

2

1

,

4

5

).

Normal passes through centre.

Hence we need a line which is paasing through (

2

1

,

4

5

) and (1,1)

The slope of the line is

2

1

−1

4

5

−1

=−

2

1

⇒ Equation of normal is: (y−1)=−

2

1

(x−1)

⇒2y−2=−x+1

⇒x+2y−3=0