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Exercise 10.1 · Q15

Q.Does the point (−2.5,3.5)(-2.5, 3.5) lie inside, outside or on the circle x2+y2=25x^2 + y^2 = 25?

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The key idea is to compare the squared distance of the point from the origin to the squared radius. Since (−2.5)2+(3.5)2=18.5<25(-2.5)^2 + (3.5)^2 = 18.5 < 25, the point lies inside the circle.

The equation x2+y2=25x^2 + y^2 = 25 is the standard form of a circle centered at the origin (0,0)(0,0) with radius r=25=5r = \sqrt{25} = 5. For any point (x,y)(x, y), the expression x2+y2x^2 + y^2 gives the square of its distance from the center. So, to decide where a point lies relative to the circle, we compare this squared distance to r2=25r^2 = 25:

  • If x2+y2<25x^2 + y^2 < 25, the point is inside the circle.
  • If x2+y2=25x^2 + y^2 = 25, the point is on the circle.
  • If x2+y2>25x^2 + y^2 > 25, the point is outside the circle.

This works because we’re comparing distances without needing to take square roots — simpler and exact.

  1. Identify the point and the circle.

    The point is (−2.5,3.5)(-2.5, 3.5). The circle is x2+y2=25x^2 + y^2 = 25, so r2=25r^2 = 25.

  2. Compute the squared distance from the origin.

    Square each coordinate and add:

    (−2.5)2=6.25(-2.5)^2 = 6.25 and (3.5)2=12.25(3.5)^2 = 12.25.

    Sum: 6.25+12.25=18.56.25 + 12.25 = 18.5.

  3. Compare with r2r^2.

    18.5<2518.5 < 25, so the squared distance is less than the squared radius. …

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