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
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)"
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