Question

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

Answer 1

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 $Q=(5,0)$