A circle passes through the origin and has it | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(c) Let the required circle be

${x^2} + {y^2} + 2gx + 2fy + c = 0$&hellip;.$(i)$

This pases through $(0, 0)$, therfore $c = 0$.

The centre $(

Vedclass · Accepted Answer

${x^2} + {y^2} - 2x - 2y = 0$

Vedclass · Answer

(c) Let the required circle be

${x^2} + {y^2} + 2gx + 2fy + c = 0$&hellip;.$(i)$

This pases through $(0, 0)$, therfore $c = 0$.

The centre $( - g,\; - f)$ of $(i)$ lies on $y = x$, hence $g = f$.

Since $(i)$ cuts the circle ${x^2} + {y^2} - 4x - 6y + 10 = 0$ orthogonally, therefore

$2( - 2g - 3f) = c + 10 \Rightarrow - 10g = 10$

$ \Rightarrow g = f = - 1$

Hence the required circle is ${x^2} + {y^2} - 2x - 2y = 0$.

A circle passes through the origin and has its centre on $y = x$. If it cuts ${x^2} + {y^2} - 4x - 6y + 10 = 0$ orthogonally, then the equation of the circle is

Similar Questions

If the two circles, $x^2 + y^2 + 2 g_1x + 2 f_1y = 0\, \& \,x^2 + y^2 + 2 g_2x + 2 f_2y = 0$ touch each then:

If the circles $(x+1)^2+(y+2)^2=r^2$ and $x^2+y^2-4 x-4 y+4=0$ intersect at exactly two distinct points, then

Locus of the points from which perpendicular tangent can be drawn to the circle ${x^2} + {y^2} = {a^2}$, is

The coordinates of the radical centre of the three circles ${x^2} + {y^2} - 4x - 2y + 6 = 0,{x^2} + {y^2} - 2x - 4y -1 = 0,$${x^2} + {y^2} - 12x + 2y + 30 = 0$ are

Circles ${x^2} + {y^2} - 2x - 4y = 0$ and ${x^2} + {y^2} - 8y - 4 = 0$