The equation of the circle which intersects c | vedclass

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

Question

(D) Let the required circle be ${x^2} + {y^2} + 2gx + 2fy + c = 0$. Two circles ${x^2} + {y^2} + 2g_1x + 2f_1y + c_1 = 0$ and ${x^2} + {y^2} + 2g_2x +

Vedclass · Accepted Answer

$3({x^2} + {y^2}) + 4(x + y) - 3 = 0$

Vedclass · Answer

(D) Let the required circle be ${x^2} + {y^2} + 2gx + 2fy + c = 0$. Two circles ${x^2} + {y^2} + 2g_1x + 2f_1y + c_1 = 0$ and ${x^2} + {y^2} + 2g_2x + 2f_2y + c_2 = 0$ intersect at right angles if $2g_1g_2 + 2f_1f_2 = c_1 + c_2$. Applying this condition to the given circles: $(i) \; 2g(1/2) + 2f(1) = c + 3 \Rightarrow g + 2f = c + 3$ $(ii) \; 2g(1) + 2f(2) = c + 5 \Rightarrow 2g + 4f = c + 5$ $(iii) \; 2g(-7/2) + 2f(-4) = c - 9 \Rightarrow -7g - 8f = c - 9$ Subtracting $(i)$ from $(ii)$,we get $g + 2f = 2$. Substituting $g + 2f = 2$ into $(i)$,we get $2 = c + 3$,so $c = -1$. Now,$g + 2f = 2$ and $-7g - 8f = -1 - 9 = -10$,which simplifies to $7g + 8f = 10$. Solving $g + 2f = 2$ and $7g + 8f = 10$: Multiply the first by $4$: $4g + 8f = 8$. Subtracting this from $7g + 8f = 10$ gives $3g = 2$,so $g = 2/3$. Then $2/3 + 2f = 2$ $\Rightarrow 2f = 4/3$ $\Rightarrow f = 2/3$. Substituting $g=2/3, f=2/3, c=-1$ into the general equation: ${x^2} + {y^2} + 2(2/3)x + 2(2/3)y - 1 = 0$ ${x^2} + {y^2} + 4/3x + 4/3y - 1 = 0$ Multiplying by $3$: $3({x^2} + {y^2}) + 4x + 4y - 3 = 0$,which is $3({x^2} + {y^2}) + 4(x + y) - 3 = 0$.

The equation of the circle which intersects circles ${x^2} + {y^2} + x + 2y + 3 = 0$,${x^2} + {y^2} + 2x + 4y + 5 = 0$,and ${x^2} + {y^2} - 7x - 8y - 9 = 0$ at right angles is:

Similar Questions

The equation of the radical axis of the circles $2x^2 + 2y^2 - 7x = 0$ and $x^2 + y^2 - 4y - 7 = 0$ is

The equation of the circle which passes through the intersection of ${x^2} + {y^2} + 13x - 3y = 0$ and $2{x^2} + 2{y^2} + 4x - 7y - 25 = 0$ and whose centre lies on $13x + 30y = 0$ is

The equation of the radical axis of the circles $x^2 + y^2 + x - y + 2 = 0$ and $3x^2 + 3y^2 - 4x - 12 = 0$ is:

Two tangents are drawn from a point $P$ on the radical axis to two circles,touching them at $Q$ and $R$ respectively. Then,the triangle formed by joining $P, Q,$ and $R$ is:

The equation of the circle passing through the points of intersection of two circles $x^2+y^2+2x+3y+1=0$ and $x^2+y^2+4x+3y+2=0$ and the point $(-1,1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module