WEBVTT
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A radar is located at the point π΄ negative nine and negative five covering a circular region with a radius of 27 length units.
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Determine the equation of the circle that gives the boundary of the radarβs reach.
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So we have been given the coordinates of the centre of a circle and the circleβs radius and weβre asked to determine its equation.
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We have all the necessary information in order to be able to do this if we recall the centre radius form of the equation of a circle.
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If a circle has centre with coordinates β, π and radius π, then its equation is given by π₯ minus β all squared plus π¦ minus π all squared is equal to π squared.
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All we need to do is substitute the relevant values of β, π, and π.
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So letβs begin.
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β is the π₯-coordinate of the centre of the circle, so in this case, itβs negative nine.
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So we have π₯ minus negative nine all squared.
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We need to be very careful here.
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Itβs not π₯ minus nine; itβs π₯ minus negative nine.
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So be very careful with the negative signs.
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Next, we have π¦ minus π all squared.
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π is the π¦-coordinate of the centre of the circle, so itβs negative five.
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So we have π¦ minus negative five all squared.
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Then, this is equal to π squared.
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So the radius of our circle is 27.
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We have then π₯ minus negative nine all squared plus π¦ minus negative five all squared is equal to 27 squared.
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Now, thatβs the beginning of the equation of our circle, but we just need to neaten it up a little bit.
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So π₯ minus negative nine will become π₯ plus nine and π¦ minus negative five will become π¦ plus five.
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Weβll also evaluate 27 squared at this point, and that is 729.
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So here, we have the equation of the circle that gives the boundary of the radarβs reach: π₯ plus nine all squared plus π¦ plus five all squared is equal to 729.