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  • Circle(s) touching the X-axis at a distance 4 from the origin and having an intercept of length on the Y

Circle(s) touching the X-axis at a distance 4 from the origin and having an intercept of length 5\sqrt{7}on the Y-axis is/are:

 

Option: 1

One circle centered at (0,4)with radius 5\sqrt{10}


Option: 2

One circle centered at (4,0)with radius 5\sqrt{10}


Option: 3

Two circles: one centered at(0,4+5 \sqrt{10} / 2)with radius (5 \sqrt{10}-4) / 2and another centered at (0,4-5 \sqrt{10 / 2)} with radius (5 \sqrt{10}-4) / 2


Option: 4

Two circles: one centered at (4+5 \sqrt{10} / 2,0) with radius (5 \sqrt{10}-4) / 2and another centered at (4-5 \sqrt{10} 0 / 2,0) with radius(5 \sqrt{10}-4) / 2


Answers (1)

best_answer

Let the center of the circle be(a,b) where b is the distance from the origin to the X-axis, which is 4 in this case. Also, let the radius of the circle be r.

Since the circle touches the X-axis at a distance of 4 from the origin, we have|b|=4 Thus, we can assume without loss of generality that b=4

The intercept on the Y-axis iswhich is equal to5 \sqrt{10} 

which gives r=\frac{5 \sqrt{10}-4}{2}

So, the equation of the circle is (x-a)^2+(y-4)^2=r^2

(x-a)^2+(y-4)^2=\left(\frac{5 \sqrt{10}-4}{2}\right)^2

the circle intersects the Y-axis at 

(0, b \pm r)=\left(0,4 \pm \frac{5 \sqrt{10}-4}{2}\right), which gives two possible circles: Circle _1:(x-a)^2+\left(y-4+\frac{5 \sqrt{10}-4}{2}\right)^2=\left(\frac{5 \sqrt{10}-4}{2}\right)^2 Circle _2:(x-a)^2+\left(y-4-\frac{5 \sqrt{10}-4}{2}\right)^2=\left(\frac{5 \sqrt{10}-4}{2}\right)^2

Therefore, the two possible circles are centered at(a,b) and have radius\frac{5 \sqrt{10}-4}{2} and they intersect the X-axis at a distance of 4 from the origin and have an intercept of length5 \sqrt{10} on the Y-axis.

 

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