document.write( "Question 597896: In the circle whose equation is x^2 + y^2 = 49, a chord is perpendicular to a diameter at the point (0,4).\r
\n" ); document.write( "\n" ); document.write( "a. What is the length of the chord?\r
\n" ); document.write( "\n" ); document.write( "b What are the coordinates of the chords end points? \n" ); document.write( "

Algebra.Com's Answer #378382 by solver91311(24713) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "The circle is centered at the origin, so a diameter through the point (0,4) must be a vertical line, moreover it must lie in the

-axis. The perpendicular chord must then lie in a horizontal line with the equation

. Construct the chord and then construct the radii that intersect the circle at the two points of intersection of the chord and the circle. You will have created two right triangles with vertices at the origin, the point (0,4) and the points of intersection of the chord and the circle.\r
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\n" ); document.write( "\n" ); document.write( "The radius of the circle is 7, hence the measure of the hypotenuse of each of your two right triangles is also 7. Also, the measure of the short side of the triangles is 4 since the short side is on the

-axis from the origin to the point (0,4).\r
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\n" ); document.write( "\n" ); document.write( "Use Pythagoras to calculate the measure of each half of the desired chord. The measure of the cord is 2 times this value. The

-coordinates of the two endpoints of the chord are the positive and negative of this value and the

-coordinate of the endpoints is, perforce, 4.\r
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\n" ); document.write( "\n" ); document.write( "John
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\n" ); document.write( "My calculator said it, I believe it, that settles it
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