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Chord of Contact

Tangents are ...

Question

Tangents are drawn from the point $P (2, 2)$ to the circle $x^{2} + y^{2} = 1,$ touching the circle at $A$ and $B .$ Then equation of circumcircle of $△ P A B$ is

x^{2} + y^{2} + 2 x + 2 y = 0

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x^{2} + y^{2} - 2 x - 2 y = 0

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x^{2} + y^{2} + 2 x - 2 y = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y^{2} - 2 x + 2 y = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

The correct option is B $x^{2} + y^{2} - 2 x - 2 y = 0$

For the required circle, $P (2, 2)$ and $(0, 0)$ will be end points of its diameter.
Hence, equation will be
$(x - 2) (x - 0) + (y - 0) (y - 2) = 0 \Rightarrow x^{2} + y^{2} - 2 x - 2 y = 0$

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Chord of Contact

Standard XIII Mathematics

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Tangents are drawn from the point P(2,2) to the circle x2+y2=1, touching the circle at A and B. Then equation of circumcircle of △PAB is

The correct option is B x2+y2−2x−2y=0 For the required circle, P(2,2) and (0,0) will be end points of its diameter. Hence, equation will be (x−2)(x−0)+(y−0)(y−2)=0⇒x2+y2−2x−2y=0

Tangents are drawn from the point $P (2, 2)$ to the circle $x^{2} + y^{2} = 1,$ touching the circle at $A$ and $B .$ Then equation of circumcircle of $△ P A B$ is

The correct option is B $x^{2} + y^{2} - 2 x - 2 y = 0$

For the required circle, $P (2, 2)$ and $(0, 0)$ will be end points of its diameter.
Hence, equation will be
$(x - 2) (x - 0) + (y - 0) (y - 2) = 0 \Rightarrow x^{2} + y^{2} - 2 x - 2 y = 0$