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Suppose that line segment π΄π΅ is a diameter of a circle center negative seven, four.
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If π΅ equals negative eight, six, what is the general equation of the tangent to the circle at π΄?
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Letβs sketch what we know so far.
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We have two points of this circle.
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Its center is located at negative seven, four.
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And point π΅ lies on its outside edge at negative eight, six.
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Hereβs point π΅ and the circle should look something like this.
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We also know that line segment π΄π΅ is a diameter.
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And that means line segment π΄π΅ is created by drawing a line from point π΅ through the center of the circle to point π΄.
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We are looking for the tangent to the circle at point π΄.
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The tangent line to point π΄ will be perpendicular to line segment π΄π΅.
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To find the tangent, we need a few things.
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We need to know the slope of line π΄π΅ and we need to know where point π΄ is located at.
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Letβs start here with the slope of π΄π΅.
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To find the slope of any line, weβll need two points on that line.
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We can use points π΅ and πΆ: negative eight, six; negative seven, four.
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The slope of this line equals π¦ two minus π¦ one over π₯ two minus π₯ one, four minus six over negative seven minus negative eight.
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Four minus six equals negative two.
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Negative seven minus negative eight is the same thing as negative seven plus eight.
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And that equals one.
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The slope of line π΄π΅ is negative two over one.
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We can use that information to find point π΄.
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Point π΄ is negative two down two and right one from point πΆ, down two, right one.
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Down two is the π¦-coordinate.
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Two units down from four is two and one unit to the right of negative seven is negative six.
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Point π΄ is located at negative six, two.
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We now have the slope of line π΄π΅ and point π΄.
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The tangent line to point π΄ must have a negative reciprocal slope with the slope of line π΄π΅.
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The negative reciprocal of negative two is positive one over two.
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Line segment π΄π΅ and the tangent at point π΄ are perpendicular lines.
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We now have a point and a slope for our tangent line.
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So we can use our point-slope formula: π¦ minus π¦ one equals π times π₯ minus π₯ one.
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π¦ minus two equals one-half times π₯ minus negative six.
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π₯ minus negative six can be simplified to say π₯ plus six.
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Bring everything else down.
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Now, weβre going to distribute our one-half.
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One-half times π₯ equals one-half π₯.
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One-half times six equals three.
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π¦ minus two equals one-half π₯ plus three.
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We add two to both sides of the equation.
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π¦ equals one-half π₯ plus five.
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We have one final step to do.
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The line π¦ equals one-half π₯ plus five is the tangent line to point π΄.
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But itβs not written in the general format.
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Itβs not the general equation.
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So we need to rearrange this to the general equation format.
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The general equation format is ππ₯ plus ππ¦ plus π equals zero, where π is a positive and all of the constants are integers.
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Iβll give us a little bit more space over here.
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We subtract one-half π₯ from both sides.
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Now, we have negative one-half π₯ plus π¦ equals five.
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Now, weβre going to subtract five from both sides.
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Five minus five equals zero.
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And our left side now reads negative one-half π₯ plus π¦ minus five.
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Remember I want the constant by π₯ to be a positive integer.
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So Iβm gonna multiply the whole equation by negative two.
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Negative two times negative one-half π₯ equals positive π₯.
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Negative two times π¦ equals negative two π¦.
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Negative two times negative five equals positive 10.
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Negative two times zero equals zero.
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The general equation of the tangent to the circle at point π΄ is π₯ minus two π¦ plus 10 equals zero.