(C) To find the point of contact,substitute the equation of the line $x = 0$ into the equation of the circle $x^2 + y^2 - 2x - 6y + 9 = 0$.Substitutin

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(C) To find the point of contact,substitute the equation of the line $x = 0$ into the equation of the circle $x^2 + y^2 - 2x - 6y + 9 = 0$.Substituting $x = 0$ gives:$(0)^2 + y^2 - 2(0) - 6y + 9 = 0$$y^2 - 6y + 9 = 0$This can be written as $(y - 3)^2 = 0$.Solving for $y$,we get $y = 3$.Therefore,the point of contact is $(0, 3)$.

Class 11 Mathematics 10-1.Circle and System of Circles Tangent and normal to a circle

Medium

At which point on the $y$-axis is the line $x = 0$ a tangent to the circle $x^2 + y^2 - 2x - 6y + 9 = 0$?

A
$(0, 1)$
B
$(0, 2)$
C
$(0, 3)$
D
$(0, 4)$

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10-1.Circle and System of Circles

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Tangent and normal to a circle

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At which point on the $y$-axis is the line $x = 0$ a tangent to the circle $x^2 + y^2 - 2x - 6y + 9 = 0$?

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