Question

the equation of the circle which has both the axes as its tangents and which passes through the point(1, 2) ​

Answer 1

Answer:

What is the equation of a circle which has both axes at its tangents and passes through point (1;2)?

Ad by GSK Pharma

3-in-1 protection for moms-to-be helps protect baby too.

Consult your Gynecologist for more information on 3-in-1 vaccination during pregnancy.

Learn More

2 Answers



Shambhu Bhat, Retired professor in engineering ;Very fond of mathematics

Answered June 23, 2018 · Author has 2.6K answers and 1.4M answer views

What is the equation of a circle which has both axes at its tangents and passes through point (1;2)?

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

Let x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0 be the equation of the circle.

since it touches x axis when y=0y=0

x2+2gx+c=0x2+2gx+c=0

This must be perfect square since both roots are identical for a tangent.

c=g^2.

Similarly c=f2c=f2 so f=±g.f=±g.

But (1,2)(1,2) is in first quadrant . So f=gf=g

x2+y2+2gx+2gy+g2=0(1)(1)x2+y2+2gx+2gy+g2=0

Now plug in (1,2)

1+4+2g+4g+g2=01+4+2g+4g+g2=0

g2+6g+5=0g2+6g+5=0

(g+5)(g+1)=0(g+5)(g+1)=0

g=−5g=−5 or −1−1

Plugging in (1)(1) we get

Step-by-step explanation:

plss follow me and plss mark as brainliest

Answer 2

Step-by-step explanation:

Let x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0 be the equation of the circle.

Let x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0 be the equation of the circle.since it touches x axis when y=0y=0

Let x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0 be the equation of the circle.since it touches x axis when y=0y=0 x2+2gx+c=0x2+2gx+c=0

Let x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0 be the equation of the circle.since it touches x axis when y=0y=0 x2+2gx+c=0x2+2gx+c=0 This must be perfect square since both roots are identical for a tangent.

Let x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0 be the equation of the circle.since it touches x axis when y=0y=0 x2+2gx+c=0x2+2gx+c=0 This must be perfect square since both roots are identical for a tangent.c=g^2.

Let x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0 be the equation of the circle.since it touches x axis when y=0y=0 x2+2gx+c=0x2+2gx+c=0 This must be perfect square since both roots are identical for a tangent.c=g^2.Similarly c=f2c=f2 so f=±g.f=±g.

Let x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0 be the equation of the circle.since it touches x axis when y=0y=0 x2+2gx+c=0x2+2gx+c=0 This must be perfect square since both roots are identical for a tangent.c=g^2.Similarly c=f2c=f2 so f=±g.f=±g. But (1,2)(1,2) is in first quadrant . So f=gf=g

Let x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0 be the equation of the circle.since it touches x axis when y=0y=0 x2+2gx+c=0x2+2gx+c=0 This must be perfect square since both roots are identical for a tangent.c=g^2.Similarly c=f2c=f2 so f=±g.f=±g. But (1,2)(1,2) is in first quadrant . So f=gf=g x2+y2+2gx+2gy+g2=0(1)(1)x2+y2+2gx+2gy+g2=0

Let x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0 be the equation of the circle.since it touches x axis when y=0y=0 x2+2gx+c=0x2+2gx+c=0 This must be perfect square since both roots are identical for a tangent.c=g^2.Similarly c=f2c=f2 so f=±g.f=±g. But (1,2)(1,2) is in first quadrant . So f=gf=g x2+y2+2gx+2gy+g2=0(1)(1)x2+y2+2gx+2gy+g2=0Now plug in (1,2)

Let x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0 be the equation of the circle.since it touches x axis when y=0y=0 x2+2gx+c=0x2+2gx+c=0 This must be perfect square since both roots are identical for a tangent.c=g^2.Similarly c=f2c=f2 so f=±g.f=±g. But (1,2)(1,2) is in first quadrant . So f=gf=g x2+y2+2gx+2gy+g2=0(1)(1)x2+y2+2gx+2gy+g2=0Now plug in (1,2)1+4+2g+4g+g2=01+4+2g+4g+g2=0

Let x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0 be the equation of the circle.since it touches x axis when y=0y=0 x2+2gx+c=0x2+2gx+c=0 This must be perfect square since both roots are identical for a tangent.c=g^2.Similarly c=f2c=f2 so f=±g.f=±g. But (1,2)(1,2) is in first quadrant . So f=gf=g x2+y2+2gx+2gy+g2=0(1)(1)x2+y2+2gx+2gy+g2=0Now plug in (1,2)1+4+2g+4g+g2=01+4+2g+4g+g2=0 g2+6g+5=0g2+6g+5=0

Let x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0 be the equation of the circle.since it touches x axis when y=0y=0 x2+2gx+c=0x2+2gx+c=0 This must be perfect square since both roots are identical for a tangent.c=g^2.Similarly c=f2c=f2 so f=±g.f=±g. But (1,2)(1,2) is in first quadrant . So f=gf=g x2+y2+2gx+2gy+g2=0(1)(1)x2+y2+2gx+2gy+g2=0Now plug in (1,2)1+4+2g+4g+g2=01+4+2g+4g+g2=0 g2+6g+5=0g2+6g+5=0 (g+5)(g+1)=0(g+5)(g+1)=0

Let x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0 be the equation of the circle.since it touches x axis when y=0y=0 x2+2gx+c=0x2+2gx+c=0 This must be perfect square since both roots are identical for a tangent.c=g^2.Similarly c=f2c=f2 so f=±g.f=±g. But (1,2)(1,2) is in first quadrant . So f=gf=g x2+y2+2gx+2gy+g2=0(1)(1)x2+y2+2gx+2gy+g2=0Now plug in (1,2)1+4+2g+4g+g2=01+4+2g+4g+g2=0 g2+6g+5=0g2+6g+5=0 (g+5)(g+1)=0(g+5)(g+1)=0 g=−5g=−5 or −1−1

Let x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0 be the equation of the circle.since it touches x axis when y=0y=0 x2+2gx+c=0x2+2gx+c=0 This must be perfect square since both roots are identical for a tangent.c=g^2.Similarly c=f2c=f2 so f=±g.f=±g. But (1,2)(1,2) is in first quadrant . So f=gf=g x2+y2+2gx+2gy+g2=0(1)(1)x2+y2+2gx+2gy+g2=0Now plug in (1,2)1+4+2g+4g+g2=01+4+2g+4g+g2=0 g2+6g+5=0g2+6g+5=0 (g+5)(g+1)=0(g+5)(g+1)=0 g=−5g=−5 or −1−1 Plugging in (1)(1) we get

Let x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0 be the equation of the circle.since it touches x axis when y=0y=0 x2+2gx+c=0x2+2gx+c=0 This must be perfect square since both roots are identical for a tangent.c=g^2.Similarly c=f2c=f2 so f=±g.f=±g. But (1,2)(1,2) is in first quadrant . So f=gf=g x2+y2+2gx+2gy+g2=0(1)(1)x2+y2+2gx+2gy+g2=0Now plug in (1,2)1+4+2g+4g+g2=01+4+2g+4g+g2=0 g2+6g+5=0g2+6g+5=0 (g+5)(g+1)=0(g+5)(g+1)=0 g=−5g=−5 or −1−1 Plugging in (1)(1) we getx2+y2–10x−10y+25=0x2+y2–10x−10y+25=0

Please mark me Brainliest........