System of circles practice · Easy

Option A (47) is correct because when two circles intersect orthogonally, the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ must be satisfied, and solving for k yields 47.

Option A ($\left(\pm \frac{{3}}{{2}}, 0\right)$) is correct because it represents the limiting points of the given system of circles, which are the centers of the point circles.

The correct answer is (24,26) because the point (3,4) lies inside the circle and (-3,-4) lies outside the circle, which gives the range of p.

Option A (24) is correct because the circles intersect orthogonally, and the length of the common chord can be calculated using the formula $\frac{2rR}{\sqrt{r^2+R^2}}$.

Option B ($3 x-y+4=0$) is correct because it represents the equation of the radical axis of the coaxial system of circles, given one limiting point and one circle.

Option B (50) is correct because when one circle bisects the circumference of another, the common chord passes through the center of the second circle, leading to the equation $c + d = 50$.

Option A ($2\,\,\mathrm{or} - \frac{3}{2}$) is correct because these are the values of k that satisfy the orthogonality condition for the two circles.

Option A $\left( {0,\; \pm 1} \right)$ is correct because the circle passes through the intersection points of the line $x=0$ and the given circle equation.

Option A (${x^2} + {y^2} - 16x - 18y - 4 = 0$) is correct because it represents the equation of a circle that cuts the three given circles orthogonally.

Option B ($\left( { - 2, - 1} \right)$) is correct because the center of the circle intersecting the given circles orthogonally is the radical center of the given circles.

Option B (${x^2} + {y^2} - 3y + 1 = 0$) is correct because it represents the equation of a circle passing through (1, 1) and the intersection points of the given circles.

Option B ($4 x^2+4 y^2+30 x-13 y-25=0$) is correct because it represents a circle passing through (1,1) and the intersection points of the given circles.

Option A ($x^2+y^2-2 p x \cos \alpha-2 p y \sin \alpha+2 p^2-a^2=0$) is correct because it represents the equation of a circle where the given chord is a diameter, derived using the concept of a family of circles.

Option B ($\left( {a_1^2 - a_2^2} \right) + \left( {b_1^2 - b_2^2} \right) = r_1^2 - r_2^2$) is correct because it is derived from the condition that the common tangent passes through the origin.

Option D is correct because the locus of the centers is found by using the orthogonality condition of circles, resulting in the equation 4x - 6y - 13 = 0.

Option D $\left( {1,\; - 4} \right)$ is correct because it satisfies the equation of the common chord obtained by subtracting the equations of the two circles.

Option C (5) is correct because the number of radical axes ($^nC_2$) equals the number of radical centers ($^nC_3$) when $n=5$.

Option C ($(-1,1)(1 / 5,8 / 5)$) is correct because these are the limiting points of the coaxial system containing the two given circles.

Option C (3) is correct because the radius is calculated using the orthogonality condition for circles: $(C_1C_2)^2 = r_1^2 + r_2^2$, where $C_1$ and $C_2$ are the centers of the circles.

Option C (${x^2} + {y^2} - 6x - 6y + 9 = 0$) is correct because it satisfies the orthogonality condition, passes through (3, 0), and touches the y-axis.

Option D (${y^2} = 16x$) is correct because it represents the locus of the center of the circle that cuts the given circle orthogonally and touches the line x=2, derived by eliminating the constant term and finding the relationship between g and f.

Option D (-3, 6) is correct because it satisfies the equation of the circle formed by the intersection of the given circles and having its center on the given line.

Option B (orthocentre of the triangle) is correct because the common chords are the altitudes of the triangle, which intersect at the orthocenter.

Option A (3) is correct because the radius is calculated using the condition for orthogonal intersection of two circles.

Option B ($8 x-12 y+5=0$) is correct because it represents the locus of the center of the circle that cuts the given circles orthogonally, derived from the orthogonality condition.

Option D ($x-y-1=0$) is correct because it represents the equation obtained by subtracting the equations of the two circles, which gives the common chord.

Option D $\left( {1,\; - 4} \right)$ is correct because it is the only point that satisfies the equation of the common chord, which is found by subtracting the equations of the two circles.

Option A ($x = 3$) is correct because the equation of the common tangent is found by subtracting the equations of the two circles, ${S_1} - {S_2} = 0$.

Option C ($aa' + bb' = \frac{c}{2} + c'$) is correct because the condition for orthogonal intersection of two circles ($2(g_1g_2 + f_1f_2) = c_1 + c_2$) is applied after adjusting the second equation.

Option C ($- \frac{{10}}{3}$) is correct because it is the value of k that satisfies the condition for two circles to cut orthogonally.