# Video: Finding the General Equation of a Line Tangent to a Circle

Suppose that line segment 𝐴𝐵 is a diameter of a circle center (−7, 4). If 𝐵(−8, 6), what is the general equation of the tangent to the circle at 𝐴?

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### Video Transcript

Suppose that line segment 𝐴𝐵 is a diameter of a circle center negative seven, four. If 𝐵 equals negative eight, six, what is the general equation of the tangent to the circle at 𝐴?

Let’s sketch what we know so far. We have two points of this circle. Its center is located at negative seven, four. And point 𝐵 lies on its outside edge at negative eight, six. Here’s point 𝐵 and the circle should look something like this. We also know that line segment 𝐴𝐵 is a diameter. And that means line segment 𝐴𝐵 is created by drawing a line from point 𝐵 through the center of the circle to point 𝐴.

We are looking for the tangent to the circle at point 𝐴. The tangent line to point 𝐴 will be perpendicular to line segment 𝐴𝐵. To find the tangent, we need a few things. We need to know the slope of line 𝐴𝐵 and we need to know where point 𝐴 is located at.

Let’s start here with the slope of 𝐴𝐵. To find the slope of any line, we’ll need two points on that line. We can use points 𝐵 and 𝐶: negative eight, six; negative seven, four. The slope of this line equals 𝑦 two minus 𝑦 one over 𝑥 two minus 𝑥 one, four minus six over negative seven minus negative eight. Four minus six equals negative two. Negative seven minus negative eight is the same thing as negative seven plus eight. And that equals one. The slope of line 𝐴𝐵 is negative two over one.

We can use that information to find point 𝐴. Point 𝐴 is negative two down two and right one from point 𝐶, down two, right one. Down two is the 𝑦-coordinate. Two units down from four is two and one unit to the right of negative seven is negative six. Point 𝐴 is located at negative six, two.

We now have the slope of line 𝐴𝐵 and point 𝐴. The tangent line to point 𝐴 must have a negative reciprocal slope with the slope of line 𝐴𝐵. The negative reciprocal of negative two is positive one over two. Line segment 𝐴𝐵 and the tangent at point 𝐴 are perpendicular lines.

We now have a point and a slope for our tangent line. So we can use our point-slope formula: 𝑦 minus 𝑦 one equals 𝑚 times 𝑥 minus 𝑥 one. 𝑦 minus two equals one-half times 𝑥 minus negative six. 𝑥 minus negative six can be simplified to say 𝑥 plus six. Bring everything else down. Now, we’re going to distribute our one-half. One-half times 𝑥 equals one-half 𝑥. One-half times six equals three. 𝑦 minus two equals one-half 𝑥 plus three. We add two to both sides of the equation. 𝑦 equals one-half 𝑥 plus five.

We have one final step to do. The line 𝑦 equals one-half 𝑥 plus five is the tangent line to point 𝐴. But it’s not written in the general format. It’s not the general equation. So we need to rearrange this to the general equation format. The general equation format is 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 equals zero, where 𝑎 is a positive and all of the constants are integers.

I’ll give us a little bit more space over here. We subtract one-half 𝑥 from both sides. Now, we have negative one-half 𝑥 plus 𝑦 equals five. Now, we’re going to subtract five from both sides. Five minus five equals zero. And our left side now reads negative one-half 𝑥 plus 𝑦 minus five.

Remember I want the constant by 𝑥 to be a positive integer. So I’m gonna multiply the whole equation by negative two. Negative two times negative one-half 𝑥 equals positive 𝑥. Negative two times 𝑦 equals negative two 𝑦. Negative two times negative five equals positive 10. Negative two times zero equals zero.

The general equation of the tangent to the circle at point 𝐴 is 𝑥 minus two 𝑦 plus 10 equals zero.