Answer to Question #293446 in Geometry for Njm

Question #293446

The equation of a circle is given by 2x^2+2y^2-8x+5y-10=0 find the coordinates of P and Q if the circle cuts the x-axis as at the points P and Q

1
Expert's answer
2022-02-04T05:28:20-0500

Equation of the circle is, "2x^2+2y^2-8x+5y-10=0". We first write this equation in the form, "(x - h)^ 2 + (y - k) ^2 = r ^2" where "(h,k)" and "r" are the center and radius of the circle.

So,

"2x^2+2y^2-8x+5y-10=0\\implies 2x^2+2y^2-8x+5y=10". Putting the "x" terms and "y" terms together,

"2x^2-8x+2y^2+5y=10"

Dividing through by 2,

"x^2-4x+y^2+{5\\over2}y=5"

Completing the squares,

"(x^2-4x+4)+(y^2+{5\\over2}y+{25\\over16})=5+4+{25\\over16}={169\\over16}"

We can rewrite this equation as,

"(x-2)^2+(y+{5\\over4})^2={169\\over16}...........(1)".This is the equation of the circle.

The center of this circle has coordinates "(2,-{5\\over 4})" and radius "r={13\\over4}"


To find the "x" intercept, we set "y=0" in equation 1.

Setting "y=0" in equation 1,

"(x-2)^2+({5\\over4})^2={169\\over16}\\implies (x-2)^2={144\\over16}\\implies x-2=\\pm\\sqrt{144\\over16}"

The values of "x" when "y=0" are therefore,

"x-2=\\pm\\sqrt{144\\over16}\\implies x=2\\pm3"

Therefore,

"x=-1,5"

Thus, the "x" intercepts are, "P=(-1,0)" and "Q=(5,0)"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS