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  • Let C be the circle with the centre at (2, 3) and pass through the point (5, 7). If P is a point on the circle such that the line passing through (2, 3) and (P.x, P.y) has a slope of 2, then the co

Let C be the circle with the centre at (2, 3) and pass through the point (5, 7). If P is a point on the circle such that the line passing through (2, 3) and (P.x, P.y) has a slope of 2, then the coordinates of P are:

 

Option: 1

(5, 9)

 


Option: 2

(1, 1)


Option: 3

(3, 7)

 


Option: 4

(4, 8)


Answers (1)

best_answer

Let the coordinates of P be \mathrm{\left ( x,y \right )}. Since P lies on the circle with the centre at (2, 3) and passing through (5, 7), we have:

(x-2)^2+(y-3)^2=(5-2)^2+(7-3)^2

Simplifying, we get:

(x-2)^2+(y-3)^2=29

Also, the line passing through \left ( 2,3 \right ) and \mathrm{\left ( x,y \right )} has a slope of 2. Therefore, we have:

\frac{(y-3)}{(x-2)}=2

Simplifying, we get:

y=2 x-1

Substituting this in the equation of the circle, we get:

(x-2)^2+(2 x-4)^2=29

Simplifying, we get a quadratic equation:

5 x^2-16 x+11=0

Solving this equation, we get x=1 \text { or } \frac{11}{5} Since P lies in the circle, we must have:

(y-3)^2=29-(x-2)^2

Substituting \mathrm{x=1}, we get \mathrm{y=1}, does not satisfy the equation of the line. Therefore, we must have \mathrm{x=\frac{11}{5}},and substituting this in the equation of the line, we get \mathrm{y=\frac{9}{5}}, Therefore, the coordinates of P are \left ( \frac{11}{5}, \frac{9}{5}\right ),which is closest to option (d).

 

 

 

 

 

Posted by

Devendra Khairwa

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